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.
Because you're assuming the group is compact, there's another method for classifying such spaces. This approach can be used in higher dimensions (though people often assume the resulting manifold is simply connected just to cut down on the work).
I'll also always be assuming things are connected: the components of a homogeneous space $G/H$ are always diffeomorphic, and the identity component of $G$ acts transitively on a connected component of $G/H$, so this is all ok.
Suppose $G$ is a connected compact Lie group, $H\subseteq G$ is a closed subgroup. Because $G$ is compact, we may equip the homogeneous space $G/H$ with a Riemannian metric for which the action by $G$ is isometric.
Now, consider the isotropy action of $H$ on $T_{eG} G/H$, where $e\in G$ denotes the identity element. Recall (see, e.g., this MSE question) that if an isometry of a (connected) Riemannian manifold fixes a point and acts as the identity at the tangent space at that point, then that isometry must be the identity. Elements of $h$ fix $e H\in G/H$, so, if such an $h$ acts trivially on $T_{eG}G/H$, we must have $h = e$.
Said another way, the isotropy action gives an injective map $H\rightarrow O(T_{eG} G/H)$ into the orthogonal group. Thus, we may view $H$ as a subgroup of $O(T_{eG} G/H)$.
I'll now specialize to the case where $G/H$ is $3$-dimensional, so $O(T_{eG} G/H)$ can be identified with $O(3)$. The connected subgroups of $SO(3)$ are well known: they are $\{e\}, SO(3)$, or a conjugate of the usual $SO(2)\subseteq SO(3)$. Note that $G/H$ is equivariantly diffeomorphic to $G/(gHg^{-1})$ for any $g\in G$, so, in terms of classification, we can assume the identity component $H^0$ of $H$ is given by one of $\{e\}, SO(2)$, or $SO(3)$.
We'll break into cases depending on $H^0$.
Case A
If $H^0 = \{e\}$, then $3 = \dim G - \dim H$ impies that $\dim G = 3$. From the classification of simply connected Lie groups, $G$ has a cover which is either isomorphic to $SU(2)$ or to $T^3$. In the first case $G = SU(2)$ or $SO(3)$ and $SO(3)/H = SU(2)/\pi^{-1}(H)$ with $\pi:SU(2)\rightarrow SO(3)$ the double cover, so you just get quotients of $SU(2)$. In the second case, you get quotients of $T^3$. However, $H\subseteq T^3$ is normal, so $T^3/H$ is an abelian Lie group of dimension $3$, so $T^3/H\cong T^3$ no matter what $H$ is (so long as $H^0 = \{e\}$.)
Case B
If $H^0 = SO(2)$, then $\dim G = 4$. From the classification of simply connected Lie groups, $G$ has a cover of the form $G = T^4$ or $G = SU(2)\times S^1$. If $G = T^4$, the argument from the previous paragraph establishes that $T^4/H\cong T^3$. So, we will assume that $G$ is covered by $G = SU(2)\times S^1$. In fact, we'll pull everything back: $G/H\cong (SU(2)\times S^1)/\pi^{-1}(H)$, so we'll actually work with $G= SU(2)times S^1$. (Note though, that by pulling back, $\pi^{-1}(H)$ need not act effectively on $(SU(2)\times S^1)/\pi^{-1}(H)$) any more.) Also, instead of writing $\pi^{-1}(H)$ everywhere, I'll abuse notation and just write $H$.
Lemma: Suppose $H$ acts diagonally on $G_1\times G_2$ and that the action of $H$ on $G_2$ is transitive with isotropy group $H'$. Then $(G_1\times G_2)/H$ is canonically diffeomorphic to $G_1/H'$.
Proof: Just check that the map $G_1/H'\rightarrow (G_1\times G_2)/H$ sending $g_1 H'$ to $(g_1,e)H$ is a diffeomorphism. $\square$
Using the lemma, it follows that if the projection of $H^0$ to the $S^1$ factor of $G$ is surjective, then we can write $G/H\cong SU(2)/H'$, where $H'$ is $0$-dimensional. These were classified in Case A). So, we may assume that the projection of $H^0$ to the $S^1$ factor of $G$ is trivial. That is, we may assume $H^0$ acts only on the $SU(2)$ factor of $G$, so $G/H^0\cong S^2\times S^1$.
So, we understand $H^0$ in this case, but what about $H$? Well, let's assume $H\neq H^0$, and pick $(h_1,h_2)\in H\setminus H_0$. Because the normalizer $N:=N_{SU(2)}(SO(2))$ has two components, and $h_1 \in N$, we know that $(h_1,h_2)^2$ acts as a rotation on just the $S^1$ factor of $G/H^0 \cong S^2\times S^1$. Hence, $(G/H^0)/\langle (h_1,h_2)^2\rangle \cong S^2\times S^1$, so we may as well assume that $(h_1,h_2)^2 \in H^0$, so that, in particular, $h_2 = \pm 1$. That is, we may as well assume that $H$ has precisely two components. If $h_1\in SO(2)$, then $G/H$ is diffeomorphic to $S^2\times S^1$ independent of $h_2$. If $h_1\notin SO(2)$ and $h_2 = 1$, we get $\mathbb{R}P^2\times S^1$, and if $h_1\notin SO(2)$ and $h_2 = -1$, we get the non-trivial $S^2$-bundle over $S^1$ (the quotient of $S^2\times S^1$ by the diagonal antipodal action).
This concludes case B.
Case C In this case, $H^0 = SO(3)$. Then $\dim G = 6$, so $G$ is covered by one of $T^6, SU(2)\times T^3$, or $SU(2)\times SU(2)$.
For $G=T^6$, because $H^0$ is non-abelian, there is no $H^0\subseteq G$, even up to cover.
For the other two, by replacing $H^0$ with a cover of $H^0$ (i.e., $SU(2)$), we can actually assume $G= SU(2)\times T^3$ or $G = SU(2)\times SU(2)$. However, in both cases, it is easy to see that the above Lemma applied, so we find $G/H\cong T^3/H'$ or $SU(2)/H'$ where $H'$ is $0$-dimensional, getting us back to Case A. This completes Case C.
Best Answer
Mostow, G. D., A structure theorem for homogeneous spaces, Geom. Dedicata 114, 87-102 (2005). ZBL1086.57024.
i. either $M$ itself fibers, or
ii. (The exceptional case.) Either $G$ is solvable or is semisimple and in this case $H$ has finitely many connected components. Unfortunately, this exceptional case is poorly explained in the statement of the theorem in the introduction: In this case Mostow should have allowed one more subgroup in his sequence. In Mostow's notation, $k=1$, causing appearance of a subgroup $F_{-1}$ which has to be equal to $\{1\}$, while $F_0=H$. Then the fibration in the statement of Theorem C reads: $$ H/F_{-1}=H \to G/F_{-1}=G\to G/H. $$
Then $G_k=G_1=G$, $\Gamma_k=\Gamma_1=H$ and we are in Mostow's case 3, where $G=G_1$ is semisimple or solvable and, in the semisimple case, $\Gamma_k$ has finitely many components. Mostow explains this better when he repeats the formulation of Theorem C in the end of section 4 (where this theorem is actually proven).
Case i. If a compact connected 3-manifold fibers, then either the fiber is a surface and the base is a circle or vice-versa.
(a) If the fiber is a surface, it has to be of nonnegative Euler characteristic (see part 1). Hyperbolic 3-manifolds do not admit such fibrations:
Namely, asphericity of $M$ implies that the fiber $F$ cannot be $S^2, P^2$. The long exact sequences of homotopy groups of a fibration implies that we have a short exact sequence $$ 1\to \pi_1(F)\to \pi_1(M)\to \pi_1(S^1)= {\mathbb Z}\to 1. $$ If $F$ is the torus or the Klein bottle then $\pi_1(M)$ would contain a nontrivial normal virtually abelian subgroup which is known to be impossible.
(b) On the other hand, a compact hyperbolic manifold also cannot fiber over a surface $B$ with circular fibers $F$. This again follows from asphericity of $M$ which results in a short exact sequence $$ 1\to \pi_1(F)= {\mathbb Z}\to \pi_1(M)\to \pi_1(B)\to 1. $$ Then, again we would obtain a normal infinite cyclic subgroup in $\pi_1(M)$, which is impossible.
Thus, compact hyperbolic 3-manifolds cannot admit transitive actions of Lie groups.
With more work, one can relax the compactness assumption in (2) to finiteness of volume and prove a similar theorem in higher dimensions as well.
ii. Consider now the exceptional case. If $G$ is semisimple and $H$ has finitely many components, $\pi_1(M)=\pi_1(G/H)$ is finite. This is impossible for a hyperbolic manifold. Lastly, if $G$ is solvable, so is $H$, hence, $H/H^c$ is solvable too (I am using Mostow's notation where $H^c$ denotes the identity component with respect to the Lie group topology). But then $\pi_1(M)$ is solvable, which is impossible for a hyperbolic manifold of finite volume.