Intuition behind image derivative using Fourier Transform for edges detection
In general, the time derivative property of the Fourier Transform is given as

$$\mathscr{F}[\frac{d}{dt}x(t)] = j\omega X(j\omega) $$

Notice that we can simply multiply by the frequency index in the Fourier Transform result.

For the 2D FT result:

$$\mathscr{F}[f(x,y)]= F(u,v)$$

Using the same property results in:

$$\mathscr{F}[\frac{d}{dx}f(x,y)]= uF(u,v)$$

$$\mathscr{F}[\frac{d}{dy}f(x,y)]= vF(u,v)$$

The reason this detects edges is clear if you observe how the differentiation of a 1D step function results in an impulse. Your edges approximate steps. The derivative is the slope, so where the slope is high (edges) the derivative is large.