By Riemann Existence Theorem, the surjection $f \colon \pi_1(\Sigma \setminus p_1, p_2, p_3) \to \mathbb{Z}_3$ gives a Galois cover with Galois group $\mathbb{Z}_3$ branched only at the points $p_1, p_2, p_3$.

This cover is precisely your covering $\Pi_f \colon \Sigma' \to \Sigma$. Then $\deg \Pi_f = |\mathbb{Z}_3|=3$. (Here I do not understand why you wrote that the cover is "unramified", which is false. Maybe you meant "ramified"?)

The Galois group of $\Pi_f$ acts naturally on $\Sigma'$, and the action is free outside the preimages of $p_1$, $p_2$, $p_3$. At these preimages the stabilizer is obviously $\mathbb{Z}_3$ itself, so the cover $\Pi_f$ is totally ramified over $p_1$, $p_2$, $p_3$.

Then the genus of $\Sigma'$ can be computed by Riemann-Hurwitz formula, obtaining $$2g(\Sigma')-2= |\mathbb{Z}_3| \bigg(2g(\Sigma)-2 + 3 \bigg(1-\frac{1}{3} \bigg) \bigg),$$ that is $g(\Sigma')=7$.

By Riemann Existence Theorem, the surjection $f \colon \pi_1(\Sigma \setminus p_1, p_2, p_3) \to \mathbb{Z}_3$ gives a Galois cover with Galois group $\mathbb{Z}_3$ branched only at the points $p_1, p_2, p_3$.

This cover is precisely your covering $\Pi_f \colon \Sigma' \to \Sigma$. Then $\deg \Pi_f = |\mathbb{Z}_3|=3$. (By the way, here I do not understand why you wrote that the cover is "unramified", which is false. Maybe you meant "ramified"?)

The Galois group of $\Pi_f$ acts naturally on $\Sigma'$, and the action is free outside the preimages of $p_1$, $p_2$, $p_3$. At these preimages the stabilizer is obviously $\mathbb{Z}_3$ itself, so the cover $\Pi_f$ is totally ramified over $p_1$, $p_2$, $p_3$.

Then the genus of $\Sigma'$ can be computed by Riemann-Hurwitz formula, obtaining $$2g(\Sigma')-2= |\mathbb{Z}_3| \bigg(2g(\Sigma)-2 + 3 \bigg(1-\frac{1}{3} \bigg) \bigg),$$ that is $g(\Sigma')=7$.