How exactly does the feedback stabilize the amplifier?

One way to see it is from the math: from $\hat{H}(\hat{s}) = \frac{\hat{A}}{1+\hat{A}(\hat{s}) \hat{F}(\hat{s})}$, for $\vert\hat{A}\hat{F}\vert»1$, $\vert\hat{H}\vert \sim \frac{1}{\vert\hat{F}\vert}$, which is independent of $\vert\hat{A}\vert$, so quite insensitive to any variation of the value of $\vert\hat{A}\vert$.

But here's a qualitative way of thinking about it: an op-amp gives an output voltage proportional to the difference between its inputs, by a large amplification factor. What the feedback network does is to send back a fraction of the output to the input. The amplifier then sees a smaller difference between the inputs... so it adjusts the output to be smaller. A fraction of this smaller output then gets sent back to the inputs again, and once again the output adjusts... this process keeps happening until the fractional voltage fed back from the output gives a difference between the inputs that no longer results in a change in the output. If the fed-back voltage at the input, is, say, $F= 0.1$ of the output, then the actual voltage at the output is $V_{\rm out}=1$ V for an input difference of $V_{\rm in}=0.1$ V, so the gain is $G=1/F=10$. For a real op-amp this all happens very fast and it all comes to equilibrium very quickly.