Why is Frost's Circle an extension of Huckel's 4n+2 rule? What physics leads to the emergent property that the circle can predict relative MOs?
The MO energies are obtained from Hückel theory: see Pi molecular orbitals of polyenes (and the Internet in general, there are many excellent expositions) for more details.

Regarding the Frost circles, they do not actually have any meaning: they are simply visual representations of the energies obtained from Hückel theory for cyclic polyenes. The last line of porphyrin's answer is relevant here: 


  In the case of circular polyenes the solutions are $E_j=\alpha + 2\beta \cos(2\pi j/n)$.


where $n$ is the number of p-orbitals (e.g. for benzene $n = 6$) and $j$ is an integer satisfying $1 \leq j \leq n$. (The convention $0 \leq j \leq n-1$ is also sometimes seen, but it is mathematically equivalent.)

I am not a mathematician so some of the notation may be less than ideal, but the idea is as follows. Note that if you draw a circle with radius $r$, then the coordinates of any point on the circle $P$ may be described as $(r\sin\theta, -r\cos\theta)$ where $\theta$ is the (clockwise) angle that $\overrightarrow{OP}$ makes with the negative $y$-axis $\overrightarrow{OY}$. For example, here is a circle with $r = 2$. The angle $\theta$ here is equal to $\pi/4$, so the coordinates of the point $P$ are given by $(2\sin(\pi/4), -2\cos(\pi/4)) = (\sqrt{2}, -\sqrt{2})$.



Now we simply need to choose a circle with radius $r = -2\beta$ (note that $\beta &lt; 0$, hence we need the negative sign) and select the six points on the circle where $\theta = 2\pi j/n$. The $y$-coordinates of these six points will then be the six MO energies. I have illustrated this for $n = 6$ and $\beta = 1$. We need to pick the points with

$$\theta = \frac{\pi}{3}, \frac{2\pi}{3}, \pi, \frac{4\pi}{3}, \frac{5\pi}{3}, 2\pi$$

and these are $A$, $B$, $C$, $D$, $E$, and $F$ in order. Note that $\angle YOA = \pi/3$, $\angle YOB = 2\pi/3$, and so on. And note that whatever the value of $n$, there will always be the case where $j = n$ and hence $\theta = 2\pi$: that corresponds to the vertex at the bottom of the circle.