The electric field a few wavelengths away from the aperture is given by the Fresnel transform (which is an approximation of the Fresnel-Kirchhoff equation).
The Fresnel Transform is the Fraunhofer Transform with a quadratic phase term in the integral:
$$ U(x,y) = \frac{e^{jkz}}{j\lambda z}{e^{k \frac{k}{2z} (x^2 + y^2) }} \int \int_{-\infty}^{\infty} \{ U \left( \xi, \eta \right) e^{j \frac{k}{2z} (\xi^2 + \eta^2 ) } \} e^{-j \frac{2 \pi}{ \lambda z} (x \xi + y \eta ) } d \xi d\eta $$
You see the quadratic phase inside the integral: $ e^{j \frac{k}{2z} (\xi^2 + \eta^2 ) } $
After a certain distance, $z$ is so large that the quadratic phase is approximately $ \approx 1 $. Since it is in the denominator of the argument of the quadratic phase.
Therefore the Fresnel Transform simplifies to the Fraunhofer Transform which is basically the Fourier Transform.
However, in a very special configuration using a lens, it just so happens that the lens imparts a quadratic phase that is equal and opposite to the quadratic phase inside the integral and they cancel out.
So by going through the lens and propagating to the rear focal plane, the quadratic phase term in the Fresnel Transform is canceled out and the Fresnel Transform is simplified to the Fraunhofer transform. And the Fraunhofer transform is the Fourier Transform.
The full answer would require several pages of math and pictures but that's basically the idea of it.