Why does the Born approximation for the scattering amplitude depend on the potential $V$ everywhere in space, unlike classical scattering?

The answer, given by @user196574 is exshaustive, I only want to add, that integration over all space arises solely due to the choice of basis solution - a plane wave, with a fixed momentum, which exists everywhere in space. If you want to consider a scattering of localized object, for example, a Gaussian wavepacket, then performing first integral over momenta $k$: $$ int d^3 k int d^3 r e^{-i (k - k_0) r} V(r) e^{-alpha k^2 / 2} = int d^3 r V(r) e^{-(r-r_0)^2 / 2 alpha} $$ you have an expression, localized in space (decaying fast with the increase of distance).

The key difference between treatments of quantum mechanical scattering and classical scattering is the nature of the incident particle. In the quantum mechanical case, the incident particle is typically treated as a delocalized plane wave, whereas in the classical case the particle is treated as a point particle. The delocalized nature of the plane wave is what leads to an integral over all space.

Of course, the quantum treatment with a delocalized plane wave is just a choice of incident wave function that is useful. In particular, because the Schrodinger equation is linear, we can represent some more localized incident particle as a superposition of plane waves and simply take the appropriate linear combination of $f(theta, phi)$ to find the corresponding $f$.

Edit out of moral compunction: While I said "simply", the actual calculation will not be so simple, and one will also need to account for time-dependent phase factors if one takes superpositions of plane waves with different energies.