No. f does not matter. It has nothing to to do with f. All you did was align the bounds between the two animations so they both start and stop together. That is ok, but that doesn't solve the problem of mapping t->g(t) making the animation take 100x longer.
It would be easier to compute the values for t myself per frame and input them in
T := [seq(2^(k/10),k=0..10]
animate(plot, [fourier(f(x,t),x,w)/2^t, w= 0..100], t= T)
Now the fourier trasnform will run at full speed since it isn't trying to figure out 2^t dependence.
The point is that 2^t should not cause the fourier transform to crap out. The CAS should be able to handle it and even make the substitution of u = 2^t and take the same time to compute fourier(f(x,u),x,w);
If fourier(f(x,u),x,w) is fast then
fourier(f(x,2^t),x,w) should be fast.
This really isn't a problem with the animation, except that I'm specifically mapping t to 2^t to make the animation "smoother"(since it counters the log dependence of t in f(x,t).