When you compute the path integral of the relativistic free particle action, it's turns out to be the same as the Green's function of a classical field. This co-incidence is huge because it derives, using free-particle physics, things like the Coulomb potential, which is interaction physics.
Is there a deep reason behind this co-incidence?
That the path integral kernel is a Green's function of its associated equation of motion is a general fact, see this question and its answers and/or the Feynman/Hibbs book on path integrals.
There is no "interaction physics" in the free propagator. For instance, the fact that the Fourier transformation of a massless bosonic field produces something that looks like a Coulomb potential doesn't mean anything - the Fourier transform of the propagator does not magically produce interactions where there was none.
It is the actual interaction terms in an action that produce the connection between the Fourier transform of the propagator and the interaction potential at tree-level (see also this answer of mine). Without the interaction terms you don't have the tree-level diagram that tells you that the interaction potential to first order is the Fourier transform of the free propagator of the carrier particle.
Of course, the fact remains: To first order, the classical interaction potential associated with a force mediated by a certain field is essentially the free propagator of that field. It's a consequence of the tree-level diagram of a minimally coupled interaction being just the propagator of the carrier particle with the two interacting particles attached as external legs.
