In modelling elementary particle interactions, Feynman diagrams are used to represent the scattering amplitude which will give the crossection for the interaction.
This is a diagram for calculating the first order contribution to the elastic scattering ( taking the x axis as time, ) of an incoming e+ e- pair to an outgoing e+ e- pair. The exchanged photon is called virtual and is represented in the integral of the calculation as the photon propagator you are discussing.
The e+ and e- are real particles, because they not only have the quantum numbers of the e+ and e- but also the fixed mass, they are on mass shell. The exchanged photon is off mass shell, because the four vector of the incoming and outgoing two real particles has a non zero invariant mass, whereas photons have zero mass. In general the mass of the virtual particle enters in the denominator of the propagator, in your case the mass of the photon is zero.
My question is: does this propagator really 'propagates the photon'?
It propagates the effect of the photon within the integral which the diagram implies.
The effect of the propagator mass to the integration can be understood better with Z exchange for the same outgoing.:
The propagator of the virtual Z has the mass squared in the denominator. As the mass is close to 100GeV for low incoming energies the diagram is very suppressed with respect to photon exchange. When the invariant mass of the incoming is on the Z mass, the resonance peak is generated.
What one should keep in mind is that propagators are within integration bounds. Free particles are not. The propagator is not a real particle, on shell. It just has the quantum numbers of the particle, and its mass in the denominator, but the four vector it represents is off mass shell, and thus it is not a real particle.