What ptomato said about spatial frequencies is correct so there's not much to add there. Maybe this is a naive way of thinking about it but I made the following additional step in ptomato's explanation when I was first looking at this: The Fourier transform is just superimposing each of the plane waves (with different direction cosines $\alpha$ and $\beta$) at some plane along the propagation direction ($z=0$ in your example). The constituent plane waves interfere at this plane to give a spatial field distribution in $x$ and $y$.
One practical/real-world implementation of this comes into play when simulating 3-D linear spatial beam propagation by using the angular spectrum alongside the shift theorem. If you know the field distribution at one plane, you can Fourier transform it for the angular spectrum, multiply this by a phase shift corresponding to some arbitrary propagation distance, and inverse Fourier transform to retrieve the field after propagating, assuming that all propagation is linear in between the starting and finishing planes $z=0$ and $z\neq 0$. This is derived as follows:
Starting field in terms of angular spectral components:
1) $U(x, y, z=0)=\frac{1}{\lambda^{2}}\int_{-a}^{a}\int_{-b}^{b}\tilde{U}\left(\frac{\alpha}{\lambda},\frac{\beta}{\lambda}, z=0\right)e^{ik(\alpha x + \beta y)}d\alpha d\beta$
Here, $a$ and $b$ are the direction cosines corresponding to the plane wave components which propagate with the largest angle to the normal of the object (which is set by the Fourier grid dimensions and propagation distance in a calculation, or the detector dimensions and propagation distance in an experiment).
Propagated field in terms of angular spectral components:
2) $U(x, y, z\neq 0)=\frac{1}{\lambda^{2}}\int_{-a}^{a}\int_{-b}^{b}\tilde{U}\left(\frac{\alpha}{\lambda},\frac{\beta}{\lambda},z\neq 0\right)e^{ik(\alpha x + \beta y)}d\alpha d\beta$
$\tilde{U}\left(\frac{\alpha}{\lambda},\frac{\beta}{\lambda},z\neq 0\right)$ is unknown, but can be calculated by propagating the integrand of the first equation using the Helmholtz equation:
3) $\frac{\partial^{2}\tilde{U}}{\partial z^{2}}=-k^{2}\tilde{U}(1-\alpha^{2}-\beta^{2})$
where $\Delta z$ is the axial propagation distance. Integrating:
4) $\tilde{U}\left(\frac{\alpha}{\lambda},\frac{\beta}{\lambda},z\neq 0\right)=\tilde{U}\left(\frac{\alpha}{\lambda},\frac{\beta}{\lambda},z=0\right)e^{-ikz\sqrt{1-\alpha^{2}-\beta^{2}}}$
Substituting 4 into 2 gives the field distribution at $z$ as a function of the field distribution at $z=0$:
5) $U(x, y, z\neq 0)=\frac{1}{\lambda^{2}}\int_{-a}^{a}\int_{-b}^{b}\tilde{U}\left(\frac{\alpha}{\lambda},\frac{\beta}{\lambda},z=0\right)e^{-ikz\sqrt{1-\alpha^{2}-\beta^{2}}}e^{ik(\alpha x + \beta y)}d\alpha d\beta$
This can be much more computationally efficient than calculating $U$ pixel-by-pixel with the Fresnel integrals, or similar.
