Solution:
It is easy to observe that there are at most two acute angles in every convex $m$-gon. Moreover, if there are two acute angles then they are located at one and the same side.

Fix $l = 0, 1, \ldots, n-1$ and let $A$ and $B$ be two vertices of $M$ such that there are $l$ vertices on the arc $\widehat{AB}$. Consider the following expression
$$\binom{l}{m-2} + \binom{n}{m-2} - \binom{n-l-1}{m-2} - (l+1)\binom{n-l-1}{m-3}$$
We count in it all $m$-gons having two acute angles to the side $AB$ and all $m$-gons having one acute angle to the right of $AB$.

It is easy to see that summation on $l = 0, 1, \ldots, n-1$ and then multiplication by $2n+1$ counts every $m$-gon exactly once.

Now using the identity $\sum_{s=0}^{k} \binom{s}{t} = \binom{k+1}{t+1}$ we obtain
$$\begin{aligned} & \sum_{l=0}^{n-1} \left[ \binom{l}{m-2} + \binom{n}{m-2} - \binom{n-l-1}{m-2} - (l+1)\binom{n-l-1}{m-3} \right] \\ =\ & \sum_{l=0}^{n-1} \binom{l}{m-2} + n\binom{n}{m-2} - \sum_{s=0}^{n-1} \binom{s}{m-2} - (n-s) \sum_{s=0}^{n-1} \binom{s}{m-3} \\ =\ & n\binom{n}{m-2} - n \sum_{s=0}^{n-1} \binom{s}{m-3} + (s+1) \sum_{s=0}^{n-1} \binom{s}{m-3} - \sum_{s=0}^{n-1} \binom{s}{m-3} \\ =\ & (m-2) \sum_{s=0}^{n-1} \binom{s+1}{m-2} - \binom{n}{m-2} = (m-2)\binom{n+1}{m-1} - \binom{n}{m-2} \\ =\ & \frac{mn - 2n - 1}{m-1} \binom{n}{m-2} . \end{aligned}$$
Therefore the answer is $\frac{(2n+1)(mn-2n-1)}{m-1} \binom{n}{m-2}$.