$\newcommand{\HH}{\mathbb H}$
$\newcommand{\RR}{\mathbb R}$
$\newcommand{\ZZ}{\mathbb Z}$
$\newcommand{\SL}{\mathop {\rm SL}}$
In your question $X = G \backslash \HH$ is given as a quotient. Also,
$G$ is a finitely generated free group and a subgroup of $\SL(2,\RR)$.
As $G$ is convex co-compact it follows that $G$ is discrete. In this
case $\HH$ is the universal cover of $X$, the group $G$ is the
so-called "deck-group", and there is a non-canonical isomorphism of
$G$ with $\pi_1(X)$. (Choosing basepoints doesn't make the
isomorphism canonical.)
The last paragraph of your question doesn't make much sense. The
usual action of $\SL(2,\ZZ)$ on $\HH$ is not convex co-compact. So it
is not a good example to think about. Here is a better example which
fits your situation very tightly:
Suppose that $G$ is the free group of rank one -- that is, $G \cong
\ZZ$. Suppose that $G$ is generated by a hyperbolic isometry $\gamma$.
Let $A_\gamma \subset \HH$ be the axis of $\gamma$ acting on $\HH$.
So $A_\gamma$ is a copy of $\RR$, topologically, and $G$ acts on
$A_\gamma$ as $\ZZ$ acts on $\RR$, by translation. So the quotient $g
= G \backslash A_\gamma$ is a circle. Also $X = G \backslash \HH$ is
a hyperbolic annulus with two "flaring ends". To make this precise,
note that $g \subset X$ is an essential loop in the annulus $X$. Note
that $X - g$ has two components $L$ and $R$ (coming from the left and
right sides of $A_\gamma$ in $\HH$). $L$ is again an annulus with one
boundary on $g$. Also, $L$ has an exponentially flaring metric (as you
move away from $g$) so the other boundary of $L$ is "at infinity".
Returning to $\HH$, note that the limit set of $G$, $\Lambda_G$, is
exactly two points: the endpoints of $A_\gamma$. Let $\Omega_G =
\partial_\infty \HH - \Lambda_G$ be the domain of discontinuity of $G$.
Then $\Omega_G$ is two open sub-arcs of $\partial_\infty \HH$. As
before $G$ acts on each of these by translation. The quotients are
the circles at infinity for $L$ and $R$ respectively. These two circles
at infinity are the Gromov boundary of the hyperbolic annulus $X$.
Finally, in this example the convex core of $X = G \backslash \HH$ is
exactly the circle $g = G \backslash A_\gamma$ that we started with.
This discussion generalizes. If $G \subset \SL(2,\RR)$ is any
(torsion free) group acting convex co-compactly on $\HH$ then the
fundamental group of $X = G \backslash \HH$ is isomorphic to $G$. Also,
the Gromov boundary of $X$, $\partial_\infty X$, is homeomorphic to
$G \backslash \Omega(G)$. Each component of $\partial_\infty X$ is a
circle - there is one for each flaring annular end of $X$. All ends
are of this type - convex co-compactness rules out cusps.