If $G$ is a non degenerate symplectic group, we can decompose it as an orthogonal direct sum of hyperbolic subgroups $G_i$ (meaning non degenerated symplectic of rank 2). See for instance "Symplectic geometries over finite Abelian groups", È M Žmud', Theorem 3.8.
Now assume that $H \subset G$ is an isotropic subgroup of $G$. Can I find a decomposition into hyperbolic subgroups $G_i$ which is compatible with $H$, meaning that $H=\oplus H \cap G_i$?
The answer is no. Let's call such a $H$ a decomposable subgroup. Then clearly if $H$ is decomposable, one can find a symplectic decomposition $G=G_1 \oplus G_2$ with $G_1, G_2$ maximal isotropic orthogonal and $H=(H \cap G_1) \oplus (H \cap G_2)$. Such a $H$ is called "standard" in Locally compact abelian groups with symplectic self-duality, Amritanshu Prasad, Ilya Shapiro, M. K. Vemuri, and their Theorem 10.13 gives an example of a non standard maximal isotropic subgroup.
NB: I am still interested in the case where $G=(\mathbb{Z}/n\mathbb{Z})^{2g}$ (ie all the invariants are the same $d_i=n$), I think their counterexample uses a $G$ with different invariants $d_i$.