There are examples of free actions on $\mathbb{R}^2$ where every orbit is discrete and closed but the action is not properly discontinuous and the quotient is non-Hausdorff. The example is rather standard. I will use $G\cong {\mathbb Z}$. Let its generator act on the punctured plane $P:=\mathbb{R}^2 - \{(0,0)\}$ via
$$
(x,y)\mapsto (2 x, 2^{-1} y).
$$
This defines an action of $G$ on $P$. It is an easy exercise to verify that the action is free and every orbit is closed and discrete. But the action is not properly discontinuous. (Consider $G$-images of any compact subset $K\subset P$ whose interior has nonempty intersection with both coordinate lines.)
In order to get a similar example of action on a plane, remove from $P$ the half-line $L=\{(x,0): x< 0\}$. The resulting subset $E\subset P$ is homeomorphic to $\mathbb{R}^2$ and is $G$-invariant. The action of $G$ on $E$ again fails to be properly discontinuous. Moreover, the quotient $E/G$ is non-Hausdorff, hence, is not a manifold (in the usual sense). However, the quotient map $E\to E/G$ is a covering map.
One can modify this example so that the quotient by $G$ is not a covering map. Namely, in $P$ consider the positive quadrant
$$
Q=\{(x,y): x> 0, y> 0\}.
$$
Next, take $T:= P \setminus Q$. Define $F$ to be the quotient space of $T$ where we identity boundary arcs via $(x,0)\sim (0,y)$ ($x>0, y>0$) via the map $(x,0)\mapsto (0, x^{-1})$. The action of $G$ on $P$ preserves $T$ and descends to the quotient space $F= T/\sim$. It is easy to check that every $G$-orbit in $F$ is discrete, closed, but the map $F\to F/G$ is not a covering map (say, at the equivalence class of (0,1)). In this example, $F$ is homeomorphic to the open Moebius band. To obtain an example of an action on a simply-connected space, remove from $F$ the projection of the half-line $L$.
More interestingly, there are examples of smooth free actions of ${\mathbb R}$ on $\mathbb{R}^2$ such that the quotient is Hausdorff, each orbit is closed but the action is not proper. The quotient is homeomorphic to $[0,1)$.
Lastly, in the cited book (Vinberg et al) they deal with isometric group actions; for those it is easy to prove that every free isometric action with discrete and closed orbits (on a metric space) is properly discontinuous, hence, a covering action. You do not even need local compactness of the space. Few years ago I wrote a note discussing issues related to different definitions of proper discontinuity (here).
Best Answer
I think the answer to the first question is yes and the answer to the second one is no:
Yes, the quotient is an orbifold. The action of the finite group $G_x$ in a neighbourhood of $x$ can be linearized (at least if the action is by diffeomorphisms, I don't know about $C^0$ regularity), and the quotient $M/G$ is locally modelled on $G_x \backslash T_xM / T_x (G\cdot x)$.
No this orbifold is not good in general. For instance, you can glue a solid torus with a trivial circle fibration to a solid torus with a Seifert fibration with one singular fiber in the center and get a closed Seifert 3-manifold with one singular fiber. The fibration is given by the orbits of an action of S^1 and the quotient orbifold is a sphere with a single orbifold point, the simplest example of an orbifold not covered by a manifold.
More generally, I think every orbifold $M$ if dimension $n$ is the quotient of a manifold $P$ by an almost free action of the orthogonal group $O_n$ ($P$ is the principal $O_n$-bundle associated with the orbifold tangent bundle equipped with an orbifold Riemannian metric).