The genus of the Fermat curve $x^d+y^d+z^d=0$

When you discuss ramification points of $\pi$, the equation should be $x^d + y^d = 0$, not $x^d + y^d = 1$.

Also, whenever you wrote about $d$th roots of unity, you needed the $d$th roots of $-1$.

The rest seems right. Yes, the number of preimages of the ramification points is all that you need in order to apply Hurwitz's formula, it is not necessary to find the degree of each preimage.

Another approach to calculate the multiplicities is to use Lemma 4.4 (page 45). We calculate the multiplicities of the points $[\alpha:1:0]$ and show they are $d$.

First we need a local representation of $F$ near $[\alpha:1:0]$. In the patch where $y\neq 0$, this corresponds to $(\alpha, 0)$ in $\mathbb C^2$ (with coordinates $x, z$), cut out by $x^d+1+z^d=0$. The $z$ partial vanishes at $(\alpha, 0)$ so by implicit function theorem, the chart near $(\alpha, 0)$ is projection to the $z$ axis, and there is a holomorphic function $g(z)$ with the inverse to projection is $z \mapsto (g(z),z)$. The map $F$ on the patch $y\neq 0$ sends $(x, z)$ to $x$. So the local representation of $F$ is $z \mapsto g(z)$.

Now we need to calculate the order of vanishing of $g'$ at $0$. We have $g(z)^d+1+z^d=0$ so after differentiating and manipulating, $$ g(z)^{d-1} \cdot g'(z) = -z^{d-1}. $$ Notice $g'(0)=0$ because $g(0)=\alpha$. It follows the order of vanishing of $g'$ at $0$ is $d-1$. Hence the multiplicity is $d$.