Actually, I'm not sure the references I gave above are going to help that much. So let me just
do it here, since it's useful to have a record somewhere. The key is to have an equivariant
form the the standard Castelnuovo-de Franchis (CdF). This follows by looking at the usual proof.

Theorem. Let $M$ be a compact Kahler manifold on which a finite group $G$ acts holomorphically. Suppose that $\omega_1,\ldots \omega_n$ is a collection of linearly independent holomorphic $1$-forms such that $n>1$, $\omega_i\wedge \omega_j=0$, their span $V$ is $G$-stable. Then
there exists a $G$-equivariant holomorphic map, with connected fibres, $f:M\to C$ to a curve such that $\omega_i$ pulls back from $C$.

Proof. The statement without $G$ can be found in many places, such as Catanese, Inventiones 104 (1991). Writing $\omega_i = g_i(z)dz$ (locally), we get a map $f':M\dashrightarrow \mathbb{P}^{n-1}$
given by $f'(z) = [g_1(z),\ldots g_n(z)]$. The usual arguments show that this is defined
everywhere, the image is a curve $C'$, and that forms pullback from it. Stein factor to get
$f:M\to C$. It is clear that $G$ acts on $C$ in such a way that $f$ is equivariant.

Now suppose that $M$ is compact Kahler, and that there is a surjection $h:\pi_1(M)\to \Gamma$ onto the orbifold fundamental group of a compact hyperbolic orbifold (i.e suppose that $\Gamma$ embeds into $SL_2(\mathbb{R})$ as a cocompact properly discontinuous subgroup).

Then I claim that $h$ can be realized as by a holomorphic map of $M$
to an orbifold.

To see this, observe that
we can find a torsion free normal subgroup $\Gamma_1\subset \Gamma$ of finite index.
Let $G=\Gamma/\Gamma_1$. Let $M_1$ be the $G$-cover corresponding to $h^{-1}(\Gamma_1)$.
Let $W\subset H^1(\Gamma_1,\mathbb{C})$ be a $G$-invariant Lagrangian subspace.
If either $V=(h^*W)^{10}$ or $V=\overline{(h^*W)^{01}}$ has dimension at least $2$,
we get to appeal to CdF to get an equivariant map $M_1$ to a Riemann surface. There is
an exceptional case, which can be handled as in Catanese page 269. I won't reproduce it here.