Why are excited Slater determinants used to describe electron correlation?
It perhaps isn’t a very satisfying answer, but mathematically the reason is that this is our only option!

The atomic spin-orbitals form a complete basis set: we are guaranteed to be able to represent the true wavefunction as some linear combination of them. Under Hartree-Fock, as you point out, this is a single Slater determinant. If we want to do better than this, the extra chemistry that we want to add is electron correlation. The extra maths we need to add to cope with this is more basis elements… and the way to add these is to mix in more Slater determinants. Which is why I say this is our only option: if we start with the HF ground state, and want to mix in more basis functions still, then by necessity we will have to mix in some HF excited states.

(There is one extra point that might help in understanding the Kümmel description, which is that if we could get a lower energy for the ground state by mixing in a single-electron excitation, then the HF procedure would already have done this. So we need to consider two-electron excitations, which as Kümmel says can be interpreted as interactions between these two electrons that raise them above the HF ground state.)