Credit to Moishe Kohan https://math.stackexchange.com/a/3604136/724711:
The Uniformization Theorem says that if $S$ is a connected Riemannian surface then it is conformally equivalent to a complete Riemannian surface of constant curvature. In particular:
Every smooth (topological) connected surface $S$ (oriented or not) admits a complete metric of constant curvature. Equivalently,
$S$ is diffeomorphic (homeomorphic) to the quotient $\Gamma\backslash X$, where $X$ is either the unit sphere $ S^2 $ or the Euclidean plane $ E^2 $ or the hyperbolic plane $ H^2 $ and $\Gamma$ is a discrete subgroup of isometries of $X$ acting freely on $X$ (Note that $ S^2,E^2,H^2 $ are exactly the three simply connected symmetric spaces in dimension $ 2 $).
Part 2 does not follow from Part 1, unless you have "complete" in the statement of Part 1.
A. Let me answer some of your questions. First of all, one should be careful using Wikipedia articles as your main source. Anybody can write/edit Wikipedia pages, regardless of how little they know about the subject. In this particular case, Wikipedia's statement "Every connected two manifold admits a constant curvature metric" is correct but incomplete: The Uniformization Theorem (UT) is much stronger than the said claim. The correct claim is:
If $S$ is a connected Riemannbian surface then it is conformally equivalent to a complete Riemannian surface of constant curvature. In particular (and this is easier than the UT):
*1. Every smooth (topological) connected surface $S$ (oriented or not) admits a complete metric of constant curvature. Equivalently,
- $S$ is diffeomorphic (homeomorphic) to the quotient $\Gamma\backslash X$, where $X$ is either unit sphere or the Euclidean plane or the hyperbolic plane and $\Gamma$ is a discrete subgroup of isometries of $X$ acting freely on $X$.*
Part 2 does not follow from Part 1, unless you have "complete" in the statement of Part 1.
B. Below, I refer to $X$ from Part A as a model space. It is a basic fact that every model space is diffeomorphic to the quotient $G/K$ where $G$ is a linear Lie group and $K$ is a compact subgroup of $K$, while $G$ acts faithfully on $X=G/K$ (by left multiplication), isometrically and every isometry of $X$ comes from this action. This can be proven either by case-by-case analysis or by arguing that in dimension 2 model spaces are simply connected symmetric spaces (since each is complete and has constant curvature).
For instance, for $X={\mathbb H}^2$, $G=PO(2,1)$ (quotient of $O(2,1)$ by its center) and $K=PO(2)$ (quotient of $O(2)$ by $\pm 1$). Linearity of this group can be seen via adjoint representation of $O(2,1)$ (kernel of the adjoint representation of $O(2,1)$ is exactly the center of $O(2,1)$).
C. The projective plane does admit a Riemannian metric making it a symmetric space, namely, the quotient of the unit sphere by the antipodal map (the antipodal map is isometric with respect to the standard metric). However, $RP^2$ is not diffeomorphic to the quotient of $SU(2)$ by a subgroup.
D. Indeed, Klein bottle does admit a flat metric, but this is not because it is 2-fold covered by the 2-torus which admits a flat metric: You need a flat metric on $T^2$ invariant under a fixed-point free orientation-reversing involution. You can derive the existence of such a metric either by a direct construction or by appealing to the UT.
Best Answer
No, for dimension $ >2 $ not every locally symmetric space has constant curvature. In fact it is not even the case that every symmetric space has constant curvature. For example take $ S^2 \times S^1 $, a model for one of the eight Thurston Geometries. The $ S^2 $ is round (constant curvature $ 1 $) while the $ S^1 $ is flat (constant curvature $ 0 $). The space $ S^2 \times S^1 $ is locally symmetric (indeed it is even symmetric since it is a product of symmetric spaces). But $ S^2 \times S^1 $ cannot have constant curvature because there is no way to but the same curvature on the $ S^2 $ piece and the $ S^1 $ piece.
Although not every locally symmetric space has constant curvature, it is the case that every constant curvature space is locally symmetric. Indeed every constant curvature space has a locally isometric universal cover by $ X=G/H $ for either $ X=S^n $ if it has constant positive curvature or $ X=E^n $ if it has $ 0 $ curvature or $ X=H^n $ if it has constant negative curvature. And thus every constant curvature space can be written as a locally symmetric space as $$ \Gamma \backslash G/ H $$ where $ \Gamma $ is a discrete subgroup of isometries acting freely on $ X=G/H $ (here $ G $ is the isometry group of $ X $, and $ X $ is the simply connected symmetric space of the appropriate curvature).
And yes the canonical way to pick $ \Gamma, G, H $ for a given locally symmetric space $ M $ is to take $ X $ to be the unique simply connected symmetric space covering $ M $ then $ G $ is the isometry group of $ X $ and $ H $ is the isotropy group of the action of $ G $ on $ X $ and $ \Gamma $ is the unique subgroup of isometries acting freely on $ X $ that commutes with the universal covering map.