You need $\mu$ to be real and positive definie (i.e. $\mu$ needs to be a metric) to get an inner product on the cohomology group $H^0(X,\Omega^1)$. There is a standard construction for doing this:
We use the Hodge isomorphism theorem and harmonic forms. The metric $\mu$ on $T_X$ induces a dual metric on the holomorphic vector bundle $\Omega^1 = T_X^*$. Given two classes $[u]$ and $[v]$ in $H^0(X,\Omega^1)$, pick their $\mu$-harmonic representants $u$ and $v$, and define
$$ \langle [u], [v] \rangle = \int_X \langle u,v \rangle_\mu \, dV_\mu $$
where the inner product of $u$ and $v$ is the dual metric, and $dV_\mu$ is the volume form defined by $\mu$. The details of this construction can be found in Jean-Pierre Demailly's book on complex differential geometry, but this really is just a matter of linear algebra. An alternative source are Andrei Moroianu's notes on Kähler geometry.
In this specific case one could also proceed as follows: given $[u]$ and $[v]$ one can get an inner product by setting
$$ \langle [u], [v] \rangle = \int_X [u \wedge \overline v]. $$
One needs to check that the wedge product is well defined, but once that is done everything works because $[u \wedge \overline v]$ is a class of top degree on $X$, so we can integrate it over the curve. However, this inner product doesn't depend on the metric $\mu$.
The difference between the two constructions is that the first works for any bundle you can make out of the tangent bundle and its dual, while the second only works on cohomology groups of degree $n$ on a manifold of dimension $n$.