Sorry Ether, I'm just not being clear.
The torque available "at the motor" and "by the motor" is the same for all wheel sizes. However, keeping the same top speed and low end torque requires a different gear box (this is a common assertion by others in this thread). So the torque applied TO THE WHEEL must go through a different gearbox, and will then be a different torque. Having so designed all gearbox-wheel combinations, at stall the force at the exterior rim of any wheel will actually be the same (no losses). So the applied torque as indicated is actually F_stall * R which was used correctly.
Then I'm spinning the wheel under no load... I don't mean to set a bad example, but there is no need for a free body diagram, just the applied load (analytical dynamics). My "linear acceleration" term is also somewhat abusive, but it relates to the same setup (e.g. rad/ss converted to ft/ss). Clearly if the vehicle weighs more it will also accelerate slower. That doesn't help prove the assertion that "the vehicle accelerates slower BECAUSE the moment of inertia is higher for larger wheels." This statement was one of several independent reasons to use smaller wheels. This simple setup really lets you isolate everything.
I've shown that radius drops out of the equation entirely and that only mass and mass distribution ratio "k" contribute. In the strictest sense, I've actually disproved this assertion. YAY ME!!!
If the mass and mass distribution ratio "k" of the wheel remain constant (e.g. use a lighter material as the wheel gets larger) then the moment of inertia will actually go up (as it MUST with R squared) but there is absolutely no performance penalty!!! In fact the inertia can go up and you can increase performance using a lower mass or k value.
None of this is practical, the product of mass and k REALLY should go up in any reasonable manufacturing process. So I did not bother to argue the causality bit.
I worked out the equation because I suspected that something neat would happen to the radius, and it did.