There seems to be some confusion here.
First, yes, $H$ is a normal subgroup of $G$. But why does it follow from writing $H$ as an intersection of Stabilizer subgroups? Here's an altnerate way of thinking. By definition, a smooth group action is a homomorphism $G\rightarrow Diff(M)$. This homomorphism has a kernel and kernels are always normal. Prove that this kernel is nothing but $H$.
Second, yes $M/(G/H)$ has the structure of a smooth manifold, but it's not very interesting. The point is that the action by $G/H$ has the same orbits as the action by $G$. You've assumed your action is transitive, so $M/(G/H)$ consists of a single point - the unique connected 0-dimensional manifold.
Third, there is no reason why the action of $G/H$ on $M$ must be free. Here's a really simple example to keep in mind. Let $M = \{x,y,z\}$ consist of two points. Let $G = S_3\times \mathbb{Z}_2$ act on $M$ with the symmetric group $S_3$ acting in the usual way and the $Z_2$ factor doing nothing.
Then $H = \{e\}\times \mathbb{Z}_2$, but the action by $G/H \cong S_3$ is not free - the transpositions all fix points.
If you'd like a less dumb example, consider the action of $G = Sp(1) = \{p\in \mathbb{H}: |p| = 1\}$ on $S^2 = \{v=ai+bj+ck\in \operatorname{\mathbb{H}}| a^2 + b^2 + c^2 = 1\}$ given by $p\ast v = pvp^{-1} = pv\overline{p}$.
This action is transitive, $H = \{\pm 1\}$, but the action by $G/H$ on $S^2$ is not free. In fact, every element of $G/H$ fixes some points. For if $[p]\in G/H$ is not the identity, then $\operatorname{Im}(p)\neq 0$. Let $v = \frac{\operatorname{Im(p)}}{|\operatorname{Im(p)}|}\in S^2$. Then $p\ast v = pvp^{-1} = v$.
