Just try substituting these into the definition
F[f(x,y)](u,v)=\int^{\infty}_{-\infty}\int^{\infty}_{-\infty}f(x,y)e^{-i(ux+vy)}dxdy
f.e. let's do the first one:
f(x+6,y):
F[f(x+6,y)]=\int^{\infty}_{-\infty}\int^{\infty}_{-\infty}f(x+6,y)e^{-i(ux+vy)}dxdy
I'm going to make the next change of variables:
x->\tilde{x}-6
The boundaries of the integral remains unchanged, and so does the differential, but we get:
\int^{\infty}_{-\infty}\int^{\infty}_{-\infty}f(\tilde{x},y)e^{-i(u\tilde{x}-6u+vy)}d\tilde{x}dy
Notice that e^{i6u} is constant wrt to the integration, so we pull it out and what we have left of the integral is simply the Fourier transform of f(x,y) [because x* is just a dummy variable]:
e^{i6u}\int^{\infty}_{-\infty}\int^{\infty}_{-\infty}f(\tilde{x},y)e^{-i(u\tilde{x}+vy)}d\tilde{x}dy=e^{i6u}f(u,v)
This technique of variable change is the standard technique to observe how shift & scale of the time-domain (the original function) affects the frequency-domain (its Fourier transform).
