I think the answer to the question as it is stated is negative: if I do understand correctly the question I can prove that, if $G$ is abelian and both curves $C_i/G$ are rational, this is never true.
Indeed, since ${\mathbb P}^1$ has no torsion, $H$ is trivial, so it is enough if I construct a non trivial torsion bundle on $S$.
Let $p_1,\ldots,p_r$ be the critical values of $f_1$. These are the points whose fibre are not reduced, say of multiplicity $m_i>1$; as you know very well, since $C_1/G$ is rational and $ G $ is abelian, $r\geq 2$ and each $m_i$ is a divisor of the l.c.m. of the other $m_j$. In particular, up to renumbering the points, I can suppose that $d:=\gcd(m_1,m_2)>1$.
Set $n_i:=\frac{m_i}{d}$, call respectively $F'$ and $F''$ the reduced fibres over $p_i$ and $p_2$, so that $m_1F'$ and $m_2F''$ are fibres of $f_1$. Then the divisor $D:=n_1F'-n_2F''$ is a nontrivial d-torsion divisor. Indeed $dD=m_1F'-m_2F''=f_1^*p_1-f_1^*p_2$ is principal, and by the standard results on the normal bundles of a multiple fibre taken with the reduced structure $$lD_{|F'}=l(n_1F'-n_2F'')_{|F'}=ln_1F'_{|F'}$$ is trivial iff $ln_1$ is a multiple of $m_1$, that is iff $d$ divides $l$.
Let me add some comments. Even if we remove my assumptions, my construction, for each pair of $m_i$ not relatively prime on the same side, produces a torsion bundle, and I do not see any reason for it to be in $H$ even when $H$ is very big. Still these torsion bundle are determined by the fibrations, so I would modify your question as follows
Question: Under which assumptions is the group generated by $H$ and the bundles constructed above the whole torsion group of $S$?
That could be true in a wider range of cases. One could try to prove it in the regular+abelian case by using a theorem of Armstrong as it was used here for computing some fundamental groups.