Here's another example.
Consider $\mathbb{R}P^3$ in the model of a 3-ball with antipodal boundary points identified.
Consider the $G = \mathbb{Z}/2\mathbb{Z}$ action on $\mathbb{R}P^3$ given by the antipodal map sending $(x,y,z)$ to $-(x,y,z)$.
I claim that $\mathbb{R}P^3/G$ is homeomorphic to a cone on $\mathbb{R}P^2$. To see this, use coordinates on $C\mathbb{R}P^2$ given by $([x,y,z],t)$ where we think of $(x,y,z)\in\mathbb{R}^3$ and we're collapsing $\mathbb{R}P^2\times\{0\}$ to a point. Now,map $(x,y,z)$ in $\mathbb{R}P^3$ to $\big([x,y,z], (x^2+y^2+z^2)\big)$ (and map the origin to the cone point). This is clearly continuous away from the origin. It's not too hard to see that it's continuous at the origin as well.
It's also not hard to see that this descends to a bijective map from $\mathbb{R}P^3/G$ to $C\mathbb{R}P^2$, which is therefore a continuous bijection between compact Hausdorff spaces, so is itself a homeomorphism.
Finally, notice that $C\mathbb{R}P^2$ is not a topological manifold (with or without boundary) because of the cone point $p$. A neighborhood $U$ of the cone point $p$ has, by excision, $H_k(U, U-p)\cong H_k(C\mathbb{R}P^2, C\mathbb{R}P^2-p) = H_{k-1}(\mathbb{R}P^2)$, which means $p$ can be neither a manifold point nor a manifold-with-boundary boundary point.
One way to see this is as follows : you proved that $G/\Gamma \to H$ was a homeomorphism. $\Gamma$ is discrete so this implies that $G$ and $H$ have the same dimension.
Also, $G\to H$ is a submersion (you have to prove that), therefore on tangent spaces it is surjective, so by a dimension argument it is an isomorphism on tangent spaces, therefore $G\to H$ is a local diffeomorphism (by the local inversion theorem).
This should be enough to conclude
(By the way, you probably know this but of course the $T^n \cong \mathbb{R^n/Z^n}$ case is completely elementary and easier than the general case)
Best Answer
The map is smooth because we can construct it by passing the smooth map $G \to G_p$ via the natural quotient map $G \to G / G_p$. It follows immediately from the definitions of the two group actions that the map is equivariant. Thus, it has constant rank, and any bijective, smooth map of constant rank is a diffeomorphism. (See $\S$9 of Lee's Introduction to Smooth Manifolds (1st ed.) for more; this statement is essentially the content of Theorem 9.24 there. Possibly the relevant numbering is different in the current, second edition.)