Subgroups of index 2 in a genus-2 surface group Let $S_2$ be a compact, closed, orientable surface of genus $2$ with fundamental group $$G = \pi_1(S_2) =  \langle a_1,b_1,a_2,b_2 \mid [a_1,b_1][a_2,b_2] = 1\rangle\,.$$ Is there a method to list minimal generating sets for all subgroups of index $2$ in $G$?

What I've done:
Such subgroup must correspond to a finite covering map $\rho\colon S_3 \to S_2$ from the compact, closed and orientable surface of genus $3$. Following this answer to the question of how one may find covering maps between tori, calculating the respective generating sets for the image of the cover inside the fundamental group and considering the symmetries inherent to the solutions, I've found the following distinct subgroups:

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*$\langle a_1, b_1, a_2, b_2^2, b_2a_1b_2^{-1},b_2b_1b_2^{-1}\rangle$;

*$\langle a_1, b_1, a_2^2, b_2, a_2a_1a_2^{-1},a_2b_1a_2^{-1}\rangle$;

*$\langle a_1,b_1^2,a_2,b_2,b_1a_2b_1^{-1},b_1b_2b_1^{-1}\rangle$;

*$\langle a_1^2,b_1,a_2,b_2,a_1a_2a_1^{-1},a_1b_2a_1^{-1}\rangle$.

I know that there are $15$ such subgroups, since any surjective map $\varphi\colon G \to \mathbb{Z}/2\mathbb{Z}$ lifts to a homomorphism $\varphi'\colon F_4 \to \mathbb{Z}/2\mathbb{Z}$ of the free group on $4$ letters. The latter is in bijection with $1$-dimensional subspaces of $(\mathbb{Z}/2\mathbb{Z})^4$ over $\mathbb{Z}/4\mathbb{Z}$. This actually gives me a hint on how to build such list of generators: given any non-trivial element $w$ on $F_4/F_4^2[F_4,F_4] \simeq (\mathbb{Z}/2\mathbb{Z})^4$, I can complete it to a basis which we can always assume to contain three elements $\{x,y,z\}$ among $\{a_1,b_1,a_2,b_2\}$. If $g$ represents a lift of $w$ , the remaning problem is to find two more elements that complete the set $\{x,y,z,g^2\}$ into a generating set for the kernel of the composition $$F_4 \to \left(\mathbb{Z}/2\mathbb{Z}\right)^4 \overset{\text{projection}}{\to} \langle w \rangle\,.$$
 A: Sorry, I did a check on the computer (using Magma) and my comment was not entirely correct. It is true that the 15 subgroups are just the inverse images of the corresponding subgroups of the free groups, and so we can easily calculate generating sets with 7 elements. But their abelianizations are ${\mathbb Z}^6$, and one of their generators can be eliminated.
I tried to do one of them by hand (these days I prefer to do calculations like this by computer). The $15$ subgroups are the kernels of the $15$ nontrivial homomorphisms $G \to {\mathbb Z}/2$. As an example, let's take the subgroup to be the kernel of the homomorphism mapping $a,b,c,d$ to $1,0,1,0$, respectively (to avoid subscripts, I renamed your generators $a,b,c,d$).
This has the $7$  Schreier generators
$$b,ca^{-1},d,a^2,aba^{-1},ac,ada^{-1},$$
which I renamed $t,u,v,w,x,y,z$.
Now applying the Reidemeister-Schreier algorithm to the single group relator $aba^{-1}b^{-1}cdc^{-1}d^{-1}$ using coset representatives identity and $a$ for the subgroup produces (I hope) the two subgroup relators
$$xt^{-1}uzu^{-1}z^{-1}, wtw^{-1}x^{-1}yvy^{-1}v^{-1}.$$
You can use the first of these to eliminate one of the subgroup generators, such as $x$, resulting in a $6$-generator $1$-relator (of length $12$) presentation of the subgroup.
The calculations for the other $14$ subgroups are similar.
