Fourier Decomposition
1) A Fourier serie is written as $$\frac {a_0} 2 + \sum \limits _{m=1} ^\infty (a_m \cos \frac {2 \pi m t} T + \mathbf{i} \, b_m \sin \frac {2 \pi m t} T)$$
Unless your signal is even, i.e. $f(-t)=f(t)$ the fourier decomposition will have a real (cosine) part and an imaginary (sine) part. The middle bottom plot is more likely to be the power spectral density i.e. the fourrier transform's squared modulus. 

2) Indeed, the Fourrier basis is a orthonormal basis, in a mathematical sense. 
$$
\zeta _m(t)=\cos \frac {2 \pi m t} T + i \, \sin \frac {2 \pi m t} T
$$
$$
\int_{0}^{T}{\zeta _k(t)\zeta _l(t) \, dt}=\delta _{kl} %\quad \forall k,l\in[0,T-1]
$$

The two conclusions above still holds in the 2D case simply use as basis functions:
$$
\zeta _{u,v}(x,y)=e^{-i2\pi(ux+vy)}
$$
Define an image $f(x,y)$ and it's Fourier transform :
$$
F(u,v)=\int \int f(x,y) \zeta _{u,v} \, d x d y
$$
$F(u,v)$ isn't real unless $f(x,y)$ is even respect to $x$ and even respect to $y$, and so when building the Fourier transform's representatio, it is more likely that the represented value is the Fourier transform's squared modulus.
In the 2D case, Fourier's basis function still form a basis, it is complicated to define a base in a functional space. The two main things to note are :
First, the scalar product between basis functions $\zeta _{u_1,v_1}$ and $\zeta _{u_2,v_2}$ is always equal to 0 unless $u_1=u_2$ and $v1=v_2$ then it is equal to 1 (orthonormality). One must then prove that every function $f(x,y)$ is uniquely writtable as a linear combinaison of $\zeta _{u,v}$'s. It is true for periodic function, but I can't provide you the demonstration.

I let you check the definition of a basis for a vector space, if you understand it it will give you a good grasp of what is a basis in general.