...but I've read over and over that it's provable that a point light has 1/d2 falloff, and this formula doesn't work when the radius of the emitter is zero, so I'm still left puzzled about how to reconcile point emitters with spherical emitters.
Yes, an actual 1/d^2 falloff for arbitrarily low d would be unphysical... but so is the concept of a point light source ;). Now if you have a spherical light source you can integrate the 1/d^2 over the sphere surface and will find that it is equivalent to having a point light source at the center of the sphere as long as you are observing from outside the sphere (edit: to clarify i think you have to integrate over the whole sphere surface, so this would be a "transparent" light emitter). Also notice that while moving arbitrarily close to a theoretical point light source the energy doesn't go to infinity, the energy density does...
Generally when we speak of point light sources we mean that the radius of the light source is a lot smaller than the distance we observe from. If for example you had a linear light source then you will find a 1/d falloff and and if you have a light emitting plane there is no falloff at all (assuming the plane/line is infinite, or we are so close to it that we can consider it infinite for practical purposes)
One measures the distance as being 2 feet, and calculates a falloff factor of 1/4.
The other measures the distance as being 0.6096m, and calculates a falloff factor of 2.69.
Here you are mixing up relative and absolute measurements. You quoted "doubling the distance cuts the intensity by four"... now how do you go from doubling the distance to "2 feet"? Or more specifically, which distance did you double?
Doubling the distance would mean to go from 1 foot to 2 or from 0.3m to 0.6m in both cases you doubled the distance and quartered the intensity...