I've heard it claimed that, under suitable conditions, a symmetric space $M$ is just a quotient of Lie group $G$ by maximal compact subgroup $K$. On the other hand, I am having a tough time finding a reference for this result. For clarity, I'll add some more context now.
I am comfortable with the fact that a Riemannian manifold $M$ that is symmetric space can always be regarded as the quotient $G/G_p$, where $G:= Isom(M)$ and $G_p$ is the stabilizer of $p \in M$. So, my first question is
(1) Why under these circumstances should the stabilizer $G_p$ of any point $p$ be a maximal compact subgroup of $G$? (or what extra hypotheses are needed so that $G_p$ is maximal, compact.)
On the other hand, I have the following complementary question
(2) Given a Lie group $G$ and a maximal compact subgroup $K$, why does the quotient $G/K$ admit a Riemannian metric turning it into a symmetric space?
It's likely that I'm missing hypotheses here and there, but I'd like to know under what extra conditions we can answer these two questions.
It's only approximately true.
First, if you have a semisimple group $G$ and maximal compact subgroup $K$, then for every $G$-invariant metric $g$ on $G/K$, the Riemannian manifold $(G/K,d)$ is a symmetric space of non-compact type (= of sectional curvature $\le 0$ without Euclidean factor). The metric $g$ is not unique, but unique up to rescaling on each irreducible factor.
Symmetric spaces that are not of non-compact type cannot be written this way.
Nevertheless every symmetric space $X$ is homogeneous, and hence there exists a connected Lie group $G$ (namely $G=\mathrm{Isom}(X)^\circ$), a compact subgroup $K$, a $G$-invariant Riemannian metric $g$ on $G/K$ such that $(G/K,g)$ is isometric to $X$. Then (i) $K$ is maximal compact in $G$ if and only if one has one of the following equivalent conditions: (ii) $X$ is diffeomorphic to a Euclidean space (iii) $X$ is isometric to a product of a Euclidean space and a symmetric space of non-compact type (iv) $X$ has sectional curvature $\le 0$. This answers (1).
In (2) I guess you mean "when" and not "why":
I don't have a clear-cut answer. If $G$ is semisimple, every $G$-invariant Riemannian metric on $G/K$ is symmetric. But this also holds for $G$ a simply connected 2-dimensional Lie group ($\mathbf{R}^2$ or $\mathbf{R}\rtimes\mathbf{R}$). For a few very special higher-dimensional simply connected solvable Lie groups $G$ (namely minimal Borel subgroups of semisimple Lie groups with no compact factor) there exists some $G$-invariant Riemannian metric on $G=G/K$ that is symmetric, but I expect that it's not true for every $G$-invariant Riemannian metric in general (test-case: $\mathbf{R}^2\rtimes\mathbf{R}$ with action by homotheties, Borel in the real Lie group $\mathrm{SL}_2(\mathbf{C})$). Also the universal covering of $\mathrm{Isom}^+(\mathbf{R}^2)$ has one left-invariant Riemannian metric that is isometric to the Euclidean space $\mathbf{R}^3$, but probably not all left-invariant Riemannian metric are symmetric.
Also most 3-dimensional simply connected solvable Lie groups, say $G_t=\mathbf{R}^2\rtimes\mathbf{R}$ for action $t\mapsto\mathrm{diag}(e^t,e^{ct})$ for $c\notin\{0,1\}$, have no left-invariant Riemannian metric at all that is symmetric. The exceptions are $c=1$ for which some left-invariant Riemannian metric is isometric to the 3-dimensional real hyperbolic space $\mathbf{H}^3$, and $c=0$ for which some left-invariant Riemannian metric is isometric to $\mathbf{H}^2\times\mathbf{R}$.