Is it true that $\pi(G/H, x_0) \cong \pi(G, x_o)/ \pi(H, x_0)$? Is it true that
$$
\pi(G/H, x_0) \cong \pi(G, x_o)/ \pi(H, x_0) \text{?}
$$
if $H$ is a subgroup of $G$?
 A: If by $\pi$ you mean the fundamental group, then this is not true. Here is a counter-example:
Take $G = \mathbb{R}$ and $H = \mathbb{Z}$. The quotient $G / H$ is homeomorphic to a circle $S^{1}$ via the exponential map $t \mapsto \operatorname{exp}(2\pi it)$. But the fundamental group of the circle is non-trivial (in fact, it is isomorphic to $H$) so it can not arise as a quotient of $\pi_{1}(\mathbb{R}) = \{1\}$.
This sits in line with a more general class of counter-examples: Start with any Lie group $L$. If $L$ is not simply-connected, say $\pi_{1}(L) \cong H$, then we can associate to it the so called universal covering space. (You can read about this in the first chapter of Allen Hatcher's Algebraic Topology.) In our case, the universal covering space is a simply connected Lie group $G$ together with a covering map $p \colon G \to L$ which is at the same time a homomorphism of Lie groups. Again by the theory of covering spaces (see Hatcher) we have $\ker \ p = H$, so that $G /  H = L$. But $\pi_{1}(L) = H$ and $\pi_{1}(G) = 0$, so again $\pi_{1}(L)$ does not arise as any quotient of $\pi_{1}(G)$.
A concrete example of this would be to take $L = PSL(2,\mathbb{C})$. Then $G = SL(2,\mathbb{C})$ and the covering map $p \colon SL(2,\mathbb{C}) \to PSL(2,\mathbb{C})$ is exactly the quotient $SL(2,\mathbb{C}) \twoheadrightarrow SL(2,\mathbb{C})/\{\pm 1\} = PSL(2,\mathbb{C})$
More generally, if you have a simply connected Lie group $G$ and a Lie subgroup $H$ acting properly discontinuously on $G$ then $\pi_{1}(G/H) \cong H$... 
If you have any questions left please feel free to ask them.
