First, there is a qualitative difference between metal and glass: metal is a conductor, while glass is a dielectric. Under so called "plasma frequency" EM waves do not travel in conductors except that near the surface (see skin effect). For higher frequencies (usually far above the visible light) metal is transparent and refraction does occur. Specifically, answering to your question
What about the electron configuration of a the medium changes the net effect of the absorption and re-emissions of the photons?
Metal has a substantial amount of free electrons, glass doesn't. From the EM point of view, the difference originates from the relative effect of two terms in the Maxwell equation (roughly speaking, the corresponding dimensionless parameter depends on conductivity and wave frequency). From the QM point of view, the interaction with electron gas is substantially different from the interaction with an atom.
When you have hit the critical angle in a medium that refracts and the light completely reflects, are the photons moving is the same manner as they would be in a material that always reflects? How does this connect to the question in the previous paragraph?
Again, from the EM point of view both of these processes result in an evanescent wave and are quite similar. I am not sure what exactly happens with a specific photon (as a particle) here, as the evanescent wave is a general wave effect. Perhaps someone more knowledgeable in QED can provide a QM perspective on this.
This is one of the places where wave particle dualism gets some people in trouble. Many are taught that it means that light can be a wave and a particle, and that phrasing can lead to some confusion. I find it more intuitive to just rip the bandaid off quickly and say light is neither a wave nor a particle. It is something which, in some situations, can be well modeled as a wave, and in some situations can be well modeled as a particle, but it is its own thing (which can be well modeled in all known cases using a more complicated concept, a "wavefunction").
You can think of photons getting randomly reflected or transmitted on the boundary, but the truth is that the billiard-ball photon model really isn't very effective at describing what happens at this boundary. This is one of the regions where wave mechanics models the effects very well, while particle models don't do so well. If you use wave mechanics, the idea of a wave getting partially reflected and partially transmitted isn't difficult to believe at all. In fact, it's pretty easy to prove.
Thinking in wave terms at these boundaries also gives correct answers in peculiar situations where the particle model simply falls on its face. Consider the interesting case of an "evanescent wave."
In this setup, the laser and prism are set up at the correct angles to cause "total internal reflection." This means that, by the simple models, 100% of the light should bounce off the side of the prism and into the detector. Indeed, if the prism is in the open air, we do see 100% reflection (well, within the error bars of absorption). However, bring an object close to the prisim (but not touching) and things change. You end up seeing effects from the object, even though 100% of the light was supposed to be reflected!
If you think of light like photons, this is hard to explain. If you look at it as a wave governed by Maxwell's equations, you see that you would violate the law of conservation of energy if there was a "pure" reflection. Instead it creates a reflection and an "evanescent wave" which is outside the prism, and its strength falls off exponentially, which is really hard to explain with particles!
Of course, these too are all simplifications. The real answer to your question is that the wavefunction of the light interacts with the electromagnetic fields of the atoms in the prisim, and the result of that interaction leads to reflection, refraction, diffusion, absorption, and eveansecent waves. However, naturally those equations are a bit harder to understand, so we use the older, simpler models from before quantum mechanics. We just have to be sure to use the one which is most applicable in any given situation, because none of them are quite right.
Best Answer
the process is coherent--- the same photon is bouncing off all the atoms at once, and you only get constructive interference when the angle of reflection is equal to the angle of incidence. The condition is that the surface is smooth on the scale of the wavelength of light, so that the light can excite each atom independently, and coherently add up all their contributions. This is Feynman's explanation in QED, I think you just misunderstood it as saying that reflection is assumed--- he just assumed rescattering, and then shows you that it happens in the reflected direction preferentially.