If we take out the $2 \pi$ for simplicity's sake and rescale time such that $f_m/f_c=r$ and $f_c=1$, then we are basically considering
$$s(t)=\sin((1+\sin(rt))t) = \sin(t+t\sin(rt))$$
Now consider $r=1$ and going forward in time by $2 \pi$:
$$s(t+2 \pi)=\sin(t+2 \pi + (t+2\pi) \sin(t+2\pi)) \\\ = \sin(t + (t+2 \pi) \sin(t+2 \pi)) \\\ = \sin(t + (t + 2 \pi) \sin(t)).$$
This will be the same as $\sin(t + t \sin(t))$ provided $2 \pi \sin(t)$ is an integer multiple of $2 \pi$, which will happen for some nice cases like $t=0$ and $t=\pi/2$. But it won't be when, say, $t=\pi/6$: for $t=\pi/6$ you are effectively shifting the argument of the outer $\sin$ by $\pi$, which will change the value. Indeed $s(\pi/6)=-s(13\pi/6)$. You can see this in a plot.