How do I interpret the result of a Fourier Transform?
The Fourier transform of a function $x(t)$ is defined by

$$X(\omega)=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt\tag{1}$$

Since $e^{-j\omega t}$ is complex-valued, the Fourier transform $X(\omega)$ is generally a complex-valued function, even if $x(t)$ is real-valued.

$\omega$ is the frequency in radians per second (rad/s), and it is related to the frequency in Hertz $f$ by $\omega=2\pi f$.

Since $X(\omega)$ is generally complex-valued, it can be represented by its real and imaginary parts

$$X(\omega)=X_R(\omega)+jX_I(\omega)\tag{2}$$

or by its magnitude and phase

$$X(\omega)=|X(\omega)|e^{j\phi(\omega)}\tag{3}$$

Clearly, the real and imaginary parts are not the same as magnitude and phase.

Note that for real-valued $x(t)$, the real part $X_R(\omega)$ and the imaginary part $X_I(\omega)$ of the Fourier transform $X(\omega)$ are even and odd functions, respectively. Consequently, the negative frequency components are redundant, and knowledge of the spectrum at positive frequencies is sufficient. If you are interested in power, you should rather plot $|X(\omega)|^2=X_R^2(\omega)+X_I^2(\omega)$ (which, by definition, is non-negative, as it should be the case for energy or power). The function $|X(\omega)|^2$ shows the distribution of the frequency components of $x(t)$.

As you've noted, the DC component of the signal is zero, because its average value over time is zero:

$$X(0)=\int_{-\infty}^{\infty}x(t)dt=0\tag{4}$$