This question is from section 7, chapter 3 of An introduction to Differentiable Manifolds By William M. Boothby.
$G_{x_0}$ is the stabilizer of $x_0$. The relation that gives sense to the quotient $G/G_{x_0}$ is defined in $G$ by $g_1\sim g_2\quad\text{iff}\quad g_2=hg_1$ for some $h\in G_{x_0}$. So $G/G_{x_0}$ is defined as $G/\sim$ as usual.
The condition $\tilde{F}=F\circ\pi$ implies that if $g_1\sim g_2$ then $[g_1]=[g_2]$ and then $\begin{align*} \pi(g_1) &=\pi(g_2), &&\text{taking $F$ both sides gives}\\ \tilde{F}(g_1)&= \tilde{F}(g_2)\\ g_1x_0 &= g_2x_0\\ x_0 &= g_1^{-1}g_2x_0, \end{align*}$ that is, $g_1^{-1}g_2\in G_{x_0}$ and this implies $g_1^{-1}\sim g_2^{-1}$.
So a necessary condition for this $F$ to exist is that $G$ must satisfy $g_1\sim g_2\quad\text{implies}\quad g_1^{-1}\sim g_2^{-1}.\tag{1}\label{c}$
There are groups $G$ so that \ref{c} is not true. For such a $G$, $F$ can't exist.
My question is is it the exercise wrong? Do we really need additional conditions on $G$.
I don't see how to use that $G$ acts transitively on $X$ to solve the problem.
EDIT The convention regarding groups and subgroups acting on them is established in the previous example (7.11). Subgroups are supposed to act on the right by translations in which case everything follows as in the Ralth answer. I'll leave the question.