Here's a simpler example which I think gives some intuition for the phenomena. Consider a rigid sphere in space that is filled with an ideal gas. Now poke a hole in the sphere. How fast does the air escape? It should depend on the velocity of the air molecules compared to the size of the sphere (i.e. giving the time it would take an air molecule to traverse the sphere and have it's velocity randomized by collisions with the wall) and the size of the opening (giving the fraction of velocities which correspond to escaping). Even though in this case the pressure differential is huge, compared to a fairly small pressure differential in the case of the optical flat, if your hole is sufficiently small (it would be on the scale of 10s of nanometers wide right before optical contacting), it will clearly take a very long time for literally all the gas to escape.

When the light from surface #2 and surface #3 leave the system (travelling upwards) the two beams interfere with eachother. Whenever $2d(x,y)$ is an integer multiple of $\lambda$ there will be constructive interference and whenever $2d(x,y)$ is a half-integer multiple of lambda there will destructive interference. This interference leads to the bright and dark fringes which are seen and used to characterize the image.

Optical flatness testingppt

Apre Inst. also advertises particular interferometers for measuring thin parallel planar samples. They also have a paper describing some of the concerns in clear detail as well as a proposing a clever technique for controlling the laser source optical spectrum to get around some of the outlined issues.

2) I don't have an answer on the wedged air gap. It seems like it is just something that typically happens in the optical contacting interferometry method so it is mentioned in practical explanations of how to perform the technique.

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I have what I think is the correct answer to the second question, where I think there is are some surprisingly interesting physics. If you place an optical flat on top of another flat optic, the optical flat will indeed sit with a layer of air between it and the underlying optic. Over time, the air escapes, and gravity will optically contact the two optics. If you apply pressure, it happens faster. The question is then, why does it take so counter-intuitively long for the air to escape?

The concerns posted in the OP are legitimate and I have now found sources confirming that interferometric measurements of surfaces of thin plane-parallel optics is in fact a technically difficult task for the reasons outlined above. Here are some of the sources.

The quick answer is that the air takes longer to escape because the gap it can escape through is getting exceedingly small. For optical contacting, virtually all of the gas must escape for the contact to take, where as for normal objects, there will be surface roughness that is many orders of magnitude larger than the optical glass, allowing "contact" to happen with a much, much larger amount of air still between the objects.

The light reflected off of surface #3 had an optical path length slightly longer than that of the light reflected off of surface #2. The path length difference is a 2D function $2d(x,y)$ over the surface between the two optics where $d(x,y)$ is the distance between the two surfaces. If the optical flat is perfectly flat that $d(x,y)$ is a topographic map of the surface of the Test Piece plus possibly a tilted plane representing the wedged air gap between the two optics.**

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1b) Possibly one of the surfaces of the optical flat is AR coated so that light doesn't reflect. However Edmund optics' spec sheets indicate both surfaces of their flats are coated. Furthermore, the measurement technique doesn't specify that the secondary surface of the optic under test must be AR coated. Again, what gives?

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Opticalflat calibration

Suppose the Test Piece in this case is an uncoated parallel glass window. My understanding of the measurement is that one illuminates the setup shown in the figure with plane wave illumination from above (directed downwards so that it first passes through the optical flat and then the test piece. Now focus on surface #2 and #3. Some of this plane wave will reflect off of surface #2 and exit the way it came in (about 3.7% of incident power). Some more of this plane wave will reflect off of surface #3 and reflect the way it came in (about 3.4% of incident power)*.

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1) The discussion above only focused on the reflected light from surfaces #2 and #3 interfering. But what about light from surface #1 and #4? Basically as much light is going to come from those surfaces as from #2 and #3 and then in principle all 4 of these beams should interfere in some possibly complicated way depending on the surface deviations of each. Why is this not a consideration in all the explanations of this measurement that I see?

1) It is correct that reflections from all surfaces must be considered. The reason I hadn't come across this as a considerations in sources I consulted seems to just be that the sources I was looking at were too elementary. I think part of my surprise with this is that I would think of a plane-parallel optic as one of the most obvious sorts of optics one would want to characterize so I would have thought that the intro explanations would have explained that use case. I guess I was wrong there.

There are 40 origami lucky star strips with snowflake print against blue background. 27mm x 10mm (10 inches x .5 inches)

Opticalflat interferometer

OpticalFlat Mitutoyo

There are 40 origami lucky star strips with snowflake print against blue background. 27mm x 10mm (10 inches x .5 inches)

The total time it takes for "all" the air to escape should depend on the rate of air escaping, discussed above, the total volume of air (i.e. the size of the vacuum sphere above, or the size of the optical flat), and the exact requirement for "all" to be met. With things which are somewhat rough but very large, it can take a while as well. Imagine for example a large tarp which is droped from the ceiling. It will fall to the ground fairy fast but pockets of air will take a while to fully escape (not exactly the same processes, but similar I think).

*Actually my understanding is that at every surface interface between glass and air there is generally about 4% reflection unless there is an AR coating. My understanding is that optical flats are uncoated. This means the reflected light off of surface #2 will be 0.96*0.04*0.96 = 3.7% of the original power and the reflected light off of #3 will be 0.96*0.96*0.04*0.96*0.96 = 3.4% of the original power. Note that the reflected light off surface #1 will but 4% of the original power and the reflected light off surface #4 will be 3.1% of the total power. There are also beams which are reflected multiple times but these will all be suppressed in power by at least $0.04^2 = 0.0016$ compared to the promptly reflected beams.

2) I don't understand why there is often a wedged air gap between the two optics? If the two optics are optically contacted wouldn't there be no air between and the only deviations present would be a result of surface deviations between the two optics? Why would an air gap persist and not be close by gravity for example? In the image above it is obvious that the optical flat should just fall and close the gap between the optical flat and test piece.. why would there be a steady state wedged air gap and is this in fact desirable to make the flatness measurement more straightforward?

1a) This is barking up the right tree. Sometimes monochromatic sources are used but as the links shown here indicate, at least one of the main ways to measure plane-parallel surface optics is to use an incoherent source.

1b) This is also barking up a right tree. One of the other ways I came across to single out a particular surface is to somehow make the surfaces you don't care about non-reflective. This could be done with a coating or paint or tape or something else I guess. This seems a bit problematic because this secondary surface might be an important optical surface for your system (for example if the optic is a window) so you wouldn't want to destroy the surface in the process of making it temporarily non-reflective.

There are 40 origami lucky star strips with snowflake print against blue background.  27mm x 10mm (10 inches x .5 inches)

Optical flatness testingsoftware

1a) One possible explanation is that the coherence length of the source is shorter than the thickness of either optic meaning that you won't see interference from these other reflections, just a diffuse background reducing the contrast of the desired interference between #2 and #3. However, it seems that highly monochromatic light, such as a HeNe with a coherence length of at least meters, is used for these measurements. What gives?

4Dtechnology advertises short coherence length intereferometer for the purpose of measuring thin parallel planar optics. The fizcam 2000 datasheets indicates that the source has a coherence length of $300 \mu m$ corresponding to an incoherent source with a $\sim 1 \text{ THz}$ frequency bandwidth