No. Letting $\sigma$ and $\tau$ denote the two involutions, $\sigma \circ \tau$ is a translation by an element of $\mathrm{Pic}^0(C)$. In general, this translation will not be torsion, so its orbit through any point is infinite. (In fact, if the translation IS torsion, then $G$ is a finite dihedral group, not $\mathbb{Z}/2 \ast \mathbb{Z}/2$.

No. Letting $\sigma$ and $\tau$ denote the two involutions, $\sigma \circ \tau$ is a translation by an element of $\mathrm{Pic}^0(C)$. In general, this translation will not be torsion, so its orbit through any point is infinite. (In fact, if the translation IS torsion, then $G$ is a finite dihedral group, not $\mathbb{Z}/2 \ast \mathbb{Z}/2$.)