Spectrum of Cosine in Complex Form
The "x" axis you refer to is the frequency axis. To answer your second question first: The frequency axis is typically given as $\omega$ and waveforms as function of $\omega$ can be represented at Real and Imaginary components given by the cosines and sines, or more concisely as Magnitude and Phase as given by the complex exponential since

$$Ae^{j\omega t} = Acos(\omega t) + j Asin (\omega t) = I(t) + j Q(t)$$

Either is a representation of a complex function having real and imaginary components. For representing the frequency spectrum of a time domain function we first map our time domain function to the frequency domain with the Fourier Transform which correlates the time domain function of interest to these basis functions (either cosines and sines or much simpler the complex exponential, either with magnitude = 1). Once in the frequency domain, the result will be complex, so two plots are typically used, since it takes two real numbers to represent one complex number. This can be plots of magnitude vs frequency, and phase vs frequency, or alternatively plots of real component (I) vs Frequency and another of the imaginary component (Q) vs frequency.

When the functions are observed in the time domain (meaning how the function is changing versus time instead of versus frequency as used above) we can represent them as you suggest in terms of sine and cosine components in quadrature. This specifically is a phasor representation; for more detail on that see my explanation on the end where I use the phasor representation to help explain negative frequencies.   (There is no reason we can't use phasors to show functions in the frequency domain as well; it would be a 2 dimensional plot showing the magnitude and phase versus frequency instead of versus time- how it is plotted really comes down to what is the best way to convey the information given the function at hand). 

The basis function for the complex exponential spectrum is generally given as $e^{j\omega t}$ as given by the Fourier Transform (which is a correlation to this basis function at any given $\omega$, since correlation in general is multiply and integrate). Generally the Fourier Transform is given as: 

$$FT\{g(t)\} = G(\omega) = \int_{-\infty}^{+\infty}g(t)e^{-j\omega t}dt$$

Note since $\omega$ is a continuous function in this context (we are solving for G as a function of $\omega$), there are an infinite number of actual basis functions as the FT is an infinite-dimensional space.

DIRECT SOLUTION, k is the independent variable

If we applied that in this case, with $\omega$ and $k$ as independent variables, we get the following result:

$$FT\bigg\{\frac{1}{2}e^{jk\omega t} + \frac{1}{2}e^{-jk\omega t}\bigg\} = G(\omega)$$

$$= \frac{1}{2}\int_{-\infty}^{+\infty}e^{jk\omega t}e^{-j\omega t}dt + \frac{1}{2}\int_{-\infty}^{+\infty}e^{-jk\omega t}e^{-j\omega t}dt$$

The first term under the integral converges to an impulse (infinite height, area = 1, so 1/2 after being multiplied) when $k = +1$, as does the second term when $k = -1$. For all other k the result is zero. 

The amplitudes represent the areas of the impulses given and shows that the Fourier Transform can be given as a function of k alone in this case (since there is no dependence on $\omega$):

$$G(k) = \frac{1}{2}\delta(k+1) + \frac{1}{2}\delta(k-1)$$

We could show this as a surface plot of the magnitude for all values of $k$ and $\omega$ as shown below, with the dashed line representing infinity, which is where an impulse would occur versus k for any fixed value of $\omega$:



Thus for all $\omega$ we would obtain the following magnitude spectrum as a function of $k$ alone:



product of kw as the independent variable

If we decide to make the frequency axis be the product of $k \omega$ (since the frequency of the cosine function is given by that product), we can simply multiply the result given above by $\omega$ to result in the following magnitude spectrum:



We can similarly choose to make $\omega$ be the independent variable and would result in impulses at $\omega = \pm \frac{\omega}{k}$. All of this is in agreement that it doesn't matter what $\omega$ is, in all cases the result is an impulse in the "k domain" when k = $\pm 1$. 

Further Background on the FT of the Cosine Function

The following may help provide further intuition for those familiar with the Fourier Transform of a cosine function. 

Generally we describe the Fourier Transform of $cos(\omega t)$ as two impulses in the frequency domain $\omega$, but this applies to a specific value of $\omega$, so the time domain function of interest is given as $cos(\omega_o t)$ where $\omega_o$ represents a constant value. The Fourier Transform of $cos(\omega_o t)$ converges to two impulses in the frequency domain, but the Fourier Transform of $cos(\omega t)$ does not converge- it is infinite for all frequencies $\omega$!

What the amplitudes represent is clearest in the case where the frequency term $k\omega$ is held constant, and we define another variable such as $\Omega$ to represent all possible values of $k\omega$, as in that case the Fourier Transform (FT) of a cosine function is simply two impulses in frequency, one at a positive frequency and the other at a negative frequency.

To show this first in general, I will use capital omega $\Omega$ to avoid confusion with the $\omega$ in your formula to represent in general any angular frequency ($\Omega = 2\pi f$), and $\Omega_o$ to represent a specific frequency. 

$$FT\{\cos(\Omega_o t)\} = G(\Omega) =  \frac{1}{2}\delta(\Omega-\Omega_o) + \frac{1}{2}\delta(\Omega+\Omega_o)$$

And we can see how this relates directly to the general relationship from Euler's identity expressing the real sinusoidal cosine in terms of two complex exponential frequencies:

$$\cos(\Omega t)= \frac{1}{2}e^{j\Omega_o t} + \frac{1}{2}e^{-j\Omega_o t}$$ 

And with that we see how when we use the FT to correlate to $e^{j\Omega t}$ the result is two impulses, as when we set the frequency to either of these values $\pm \Omega_o$ the integral given by the FT above goes to infinity since $e^{j\Omega_o}e^{-j\Omega_o}=1$, but would integrate to zero anywhere else.

As a plot vs frequency this would appear as follows showing two impulses that exist at the positive and negative frequency $\Omega_o$. This is a magnitude plot, and typically to be complete a phase plot would also be shown (showing the phase versus $\omega$), but in this case it is trivial since the phase is 0 at both impulses:



Or with $k\omega$ variables, showing the magnitude spectrum that would result with both $k$ and $\omega$ as constants:



This relationship between the frequency and time waveforms as I have shown them applies for when $\Omega_o$ is a constant, otherwise we will be dealing with arbitrary modulation functions and would need to know the modulation waveform specifically (how the independent variable is changing with time) to determine the frequency spectrum. 

Negative Frequencies?

To further understand the meaning of a negative frequency, which is applicable to angular frequencies by definition, note that a negative frequency represents a single exponential frequency given as $e^{-j\omega_o t}$, with $\omega_o$ in this case any positive number. It may be helpful to visualize this with reference to the relationship I gave in the first paragraph, showing $e^{j\omega t}$ in terms of I and Q components.  To give negative frequencies a physical interpretation, we can plot the function of $e^{j\omega t}$ as a function of its real and imaginary components versus time resulting in a phasor representation, specifically the plot will map out a unit circle, and rotate around the origin at rate $\omega$ as depicted below. Here we show the angle increasing in a positive direction with time, and thus this represents a positive frequency (A negative frequency would be depicted by instead having the phasor rotate in a clockwise direction):



If this is still not clear, it may be helpful to know that the form $Ae^{j\theta}$ is identical to $A\angle \theta$

If you take the Fourier Transform of a specific exponential frequency with frequency term $-\omega_o$ given as $e^{-j\omega_o t}$, the result is a single impulse at that frequency: $\delta(\omega+\omega_o)$.  (While as we showed above the cosine function has two exponential frequencies; a positive and a negative).

The Fourier transform when presented on a graph with a positive and negative axis represents frequencies in the exponential form, so each impulse shown in the frequency domain is a $Ae^{j\omega t}$ in the time domain. When (and only when) the plot gives positive and negative frequencies where the negative frequencies are the complex conjugate of the positive frequencies given, then in that case we can also represent the same spectrum with just a positive frequency axis; as that is the only way the time domain signal can be real (using a basis function of cosine as you described); if the negative and positive frequencies are not related by a complex conjugate, then the time domain signal must be complex and a positive and negative frequency axis is required to represent the spectrum. 

This is clear from the plot below showing both $Ae^{j\omega t}$, and $Ae^{-j\omega t}$ as complex phasors rotating around the origin with time (magnitude A and angle linearly proportional with time as  $\omega t$ , which is by definition a constant frequency since $f = d\phi/ dt$).  The dotted line shows the result of summing those two phasors at any instantaneous time t, showing as long as the two phasors are related by a complex conjugate (equal magnitude opposite phase) the sum will be on the real axis and thus will be a real function with no imaginary components.  

Each spinning phasor shown is $e^{j\omega t}$ and $e^{-j\omega t}$ and their summation which stays on the real axis is a (real) cosine function.



Dilation Property of the Fourier Transform

Not specific to your question but worth mentioning since that is occuring here is the dilation property of the Fourier Transform (scaling the independent variable by a constant term k) which has the following general relationship:

$$F\{g(kt)\} = \frac{1}{|k|}G\bigg(\frac{f}{k}\bigg)$$

We could apply the above property directly by transforming the known Fourier Transform of $cos(\omega_o t)$ as another approach to arriving at a solution, although this case would be trivial to do directly as a Fourier Transform on the $cos(k\omega_o t)$ function itself.