There is an excellent expository article by Peter Scott titled "The Geometries of 3-Manifolds." The first section is spent talking about uniformization of surfaces (which is what you are interested in), and he also discusses 2-Oribfolds, which are what one gets when the group action is properly discontinuous but not free. The rest of the article is certainly good reading, but not really relevant to what you asked. Most importantly though, he sketches a proof of uniformazation for surfaces. The basic idea is that the only group which acts nicely on $S^2$ is $\mathbb{Z}/2\mathbb{Z}$, with quotient $P^2$. Furthermore, discrete groups of isometries of $\mathbb{R}^2$ with compact quotient are isomorphic to a group with a finite index subgroup isomorphic to $\mathbb{Z} \oplus \mathbb{Z}$, and hence the quotient is either a torus or a Klein bottle.Then one can classify the isometries of the hyperbolic plane, $\mathbb{H}^2$, and show that $\mathbb{Z} \oplus \mathbb{Z}$ cannot be a discrete, orientation preserving subgroup of $Isom(\mathbb{H}^2)$, and hence the torus and Klein bottle cannot admit a hyperbolic structures. Furthermore, it is easy to show (with the classification of hyperbolic isometries) that other surfaces do admit hyperbolic structures.

Also, to give you an idea of why free and properly discontinuous are important, it helps to see a few examples. First, consider $\mathbb{R}^2$ and let $G$ be the group generated by rotation around the origin through the angle $2\pi/n$. This action has a fixed point at the origin. Topologically, the quotient is just $\mathbb{R}^2$, but is has a cone point with angle $2\pi/n$ at the origin, so geometrically it is different. This type of space is what is known as an orbifold.

Without proper discontinuity, things get even worse. Consider $S^1$ and the group $G$ generated by rotation through an irrational multiple of $\pi$. The resulting quotient is not even Hausdorff (the orbit of a point under $G$ is dense in $S^1$), and we generally want to avoid that sort of thing.

In dimension 3, things get a little harder, but geometrization still works. A really good reference for this dimension is William Thurston's "3 Dimensional Geometry and Topology," which is easily one of my favorite math books ever written.

In dimension 4, the same sort of approach is a lot harder. The reason is that in dimensions 2 and 3, we have a fair deal of control as to what sorts of groups show up as fundamental groups of (closed) manifolds. It can be shown (I don't recall a reference for this off the top of my head, perhaps someone else has it) that for any finitely presented group $G$, there exists a closed 4-manifold $M$ with $\pi_1(M) = G$. So, we don't have the kind of set-up like in dimension 2 where we know which fundamental groups arise, and then show that they are isometries of one of three model spaces.

Hope this helps.