Solved! Why does the Feynman propagator model correctly the Coulomb potential 1/r?

https://journals.aps.org/pr/abstract/10.1103/PhysRev.76.769

We can read the solution from a 1949 paper by Feynman, Space-Time Approach to Quantum Electrodynamics. In Section 4 Feynman presents the Fourier decomposition of δ_+(s_21^2), which is essentially a Coulomb potential if speeds are low.

Feynman analyzes scattering using the following recipe:

For each Fourier component of 1/r, estimate the scattering of the wave function of an incoming particle.

https://en.wikipedia.org/wiki/Klein–Nishina_formula

In the derivation of the Klein-Nishina formula we see that if we have a photon field of a momentum q and an arriving electron field of a momentum p, there will be a small perturbation in the electron wave function, and the perturbation looks like a scattered (virtual) electron of a momentum p + q. We say that the scattered electron has "absorbed" a photon.

Thus, once we decompose 1/r into Fourier components, we can use Klein-Nishina like thinking to analyze scattering from each component.

The connection between Compton scattering, the Fourier decomposition of 1/r, and the Green's function for the Klein-Gordon equation

It is no coincidence that a Feynman propagator approximates (classical) Coulomb scattering well. Schrödinger's equation models classical physics well. Feynman analyzes Schrödinger's equation using a Fourier decomposition of 1/r.

But why is the Fourier decomposition of 1/r similar to the Green's function for the Klein-Gordon equation?

The reason might be that we can build a permanent source (= a static charged particle) of the electromagnetic wave equation as a sum of sources lasting an infinitesimal time. That is, the field of a static charge is an integral over Green's functions over each infinitesimal time dt.

The mundane status of a "virtual photon"

We have a partial answer to the question: what exactly is the "virtual photon" of a momentum q which charges exchange when they scatter from each other's Coulomb potential?

The answer: the virtual photon is a weird way to say that we calculate how a particle scatters from one Fourier component (whose momentum is q) of the Coulomb 1/r potential of the other particle!

That explains how the charges "know" to exchange just one photon. They do not know anything but the 1/r potential. It is our perturbation approximation which imposes this fictional exchange of a photon.

A further question: can we reduce virtual photons in all cases to a mundane Fourier decomposition of a 1/r potential? Real photons exist separate from any 1/r potential. What about virtual?