Equations for double etale covers of the hyperelliptic curve $y^2 = x^5+1$

Let $X$ be the (smooth projective model) of the hyperelliptic curve $y^2=x^5+1$ over $\mathbf C$.

Can we "easily" write down equations for all double unramified covers of $X$?

Topologically, these covers correspond to (normal) subgroups of index two in the fundamental group of $X$. So there's only a finite number of them. The function field of $X$ is $K=\mathbf C(x)[y]/(y^2-x^5-1)$. A double etale cover of $X$ corresponds to (a certain) quadratic extension $L/K$. I guess every quadratic extension of $K$ is given by taking the root of some element $D$ in $K$, or by adjoining $(1+\sqrt{D})/2$ to $K$ for some $D$.

I didn't get much further than this though.

Let me elaborate a bit my comment.

This part works for any smooth projective curve $X$ in characteristic $\ne 2$. A double cover $Y\to X$ is given, as you said, by a quadratic extension $L$ of $K={\mathbf C}(X)$. It is an elementary result that such an extension is always given by adjoining a square root $z$ of some $f\in K$: $L=K[z]$ with $z^2=f$.

Now the question is when is this cover étale ?

Claim: The cover $Y \to X$ given by $z^2=f$ is étale if and only if the divisor $\mathrm{div}_X(f)$ of $f$ is of the form $$\mathrm{div}_X(f)=2D$$ for some divisor $D$ on $X$.

It is a little too long to check all details here, but we can see that if $f$ has a zero of odd order at some point $x_0\in X$, then locally at $x_0$, we have $z^2=t^du$ where $t$ is a parameter at $x_0$, $u$ is a unit of $O_{X,x_0}$ and $d=2m+1$ is an odd positive integer. Therefore $t=(z/t^m )^2u^{-1}$ vanishes at ordre $>1$ at any point of $Y$ lying over $x_0$, so $Y\to X$ is ramified above $x_0$.

Admitting the above claim, our task now is to find rational functions $f$ on $X$ such that $\mathrm{div}(f)$ has only zeros and poles of even orders (and of course, for $L$ to be really quadratic over $K$, $f$ must not be a square in $K$). Equivalently, we are looking for divisors $D$ on $X$ such that $2D\sim 0$ and $D\not\sim 0$. A such $D$ has degree $0$ and its class in the Jacobian of $X$ has order exactly $2$.

Note that if $D'\sim D$, then the étale cover associated to $D'$ is the same then the one associated to $D$ because $\mathrm{div}(f')=2D'$ implies that $f=h^2f'$ for some $h\in K$ (the ground field needs to be algebraically closed for this), so $K[\sqrt{f}]=K[\sqrt{f'}]$. This says that $X$ has exactly $2^{2g}-1$ double étale covers, where $g$ is the genus of $X$.

If you are patient enough to arrive to this point, we can start to find the desired $D$ on a hyperelliptic curve $X$ defined by $y^2=P(x)$, with $P(x)$ of odd degree $2g+1$ ($g$ is the genus of $X$). It is known that the differences of two Weierstrass points are $2$-torsion in the Jacobian. More concretely, let $a_1, \dots, a_{2g+1}$ be the zeros of $P(X)$. Then $$f_i:=(x-a_1)/(x-a_i), \quad 2\le i\le 2g+1$$ has divisor $2(w_1-w_i)$ where $w_i$ is the Weierstrass point $x=a_i, y=0$ in $X$. If I remember correctly, the $w_1-w_i$ form a basis of the $2$-torsion points over $\mathbb Z/2\mathbb Z$. This means finally that the étale double covers of $X$ are given by $$z^2=\prod_{2\le i\le 2g+1} f_i^{\epsilon_i}, $$ for some $(\epsilon_2, \dots, \epsilon_{2g+1})\in (\mathbb Z/2\mathbb Z)^{2g}\setminus \{ 0\}$. Now you know what to do with your specific curve.

As my reputation is still too low, I could not add a comment to your answer Cantlog. What I wanted to say is the following: shouldn't it be true that the divisor of the rational function $f_{i}$ you define equals $2(w_{1} - w_{i})$, as the order of vanishing of the rational function $x$ at the point $(0,0)$ is $2$? Then indeed the class of the divisor $w_{1} - w_{i}$ is a 2-torsion point on the Jacobian of $X$, as twice that divisor becomes rationally equivalent to the class of the zero divisor.