\widehat{\mathbb C}: (z_{1},\ldots,z_{2s-1}) \mapsto-\left( \dfrac{z_{2}}{z_{1}} \right)^{2}$$ is a branched regular cover with deck group being $H_{0}$. The projection under $P$ of the set of fixed points are as follows: $$P({\rm Fix}(b_{1}))=\infty, \quad P({\rm Fix}(b_{2}))=0, \quad P({\rm Fix}(b_{3}))=1,\quad P({\rm Fix}(b_{4}))=\lambda_{1},$$ $$P({\rm Fix}(b_{2k+3}))=\mu_{k,1}, \quad   P({\rm Fix}(b_{2k+4}))=\mu_{k,2},  \quad k=1,\ldots, s.$$

In particular, the branch locus of $P$ is the set $$\{\infty,0,1,\lambda, \mu_{1,1}, \mu_{1,2},\ldots,\mu_{s-2,1}, \mu_{s-2,2}\}.$$

By the Riemann-Hurwitz formula, $D$ has genus $g_{D}=1+2^{2s-2}(s-2)$.

Let us consider the surjective homomorphism $$\theta:H_{0} \to H=\langle a_{1},\ldots,a_{s-1}\rangle \cong {\mathbb Z}_{2}^{s-1}$$ $$\begin{array}{c}
b_{1},b_{2} \mapsto a_{1}\\
b_{3},b_{4} \mapsto a_{2}\\
b_{5},b_{6} \mapsto a_{3}\\
\vdots\\
b_{2k-1}, b_{2k} \mapsto a_{k}\\
\vdots\\
b_{2s-3},b_{2s-2} \mapsto a_{s-1}\\
b_{2s-1},b_{s} \mapsto a_{s}=a_{1}a_{2}\cdots a_{s-1}
\end{array}$$

If $K$ is the kernel of $\theta$, then $$K=\langle b_{1}b_{2}, b_{3}b_{4},\ldots,b_{2k-1}b_{2k}, \ldots, b_{2s-3}b_{2s-2},b_{1}b_{3}b_{5}\cdots b_{2s-1}\rangle \cong {\mathbb Z}_{2}^{s}.$$

Since $K$ acts freely on $D$, it follows that $S=D/K$ is a Riemann surface. Again, by the Riemann-Hurwitz formula, it can be checked that the genus of $S$ is $g_{S}=1+2^{s-2}(s-2)$.

In order to write equations for $S$, we need to compute a set of generators of ${\mathbb C}[z_{1},\ldots,z_{2s-1}]^{K}$, the algebra of $K$-invariant polynomials. Since the linear action of $K$ is given by diagonal matrices, a set of generators can be found to be $$t_{1}=z_{1}^{2}, t_{2}=z_{2}^{2},\ldots, t_{2s-1}=z_{2s-1}^{2},$$ together 