A rational isotropic subspace saturates to an integral direct summand

Choose a $\mathbb{Q}$-basis $v_1,\dots ,v_g$ of $L$. Clearing the finitely many denominators in the coordinates of each $v_i$ gives nonzero integer multiples $x_i\in H\cap L$. The $x_i$ remain $\mathbb{Q}$-linearly independent, so $M=H\cap L$ has rank at least $g$; it has rank at most ${\dim }_{\mathbb{Q}}L=g$. Hence $\mathrm{rank}M=g$.

If $n\mathrm{\ne }0$, $x\in H$, and $nx\in M$, then $nx\in L$ and therefore $x=n^{-1}(nx)\in L$ because $L$ is a $\mathbb{Q}$-subspace. Thus $x\in M$, so $M$ is saturated in $H$.

We prove by induction on $r$ that every saturated subgroup $P\subset {\mathbb{Z}}^r$ is a direct summand. The cases $P=0$ and $P={\mathbb{Z}}^r$ are immediate. Otherwise choose a nonzero $p\in P$, divide its coordinates by their greatest common divisor, and use saturation to see that the resulting primitive vector $e_1$ still lies in $P$. The Euclidean algorithm gives a unimodular change of basis carrying $e_1$ to the first standard basis vector, so write ${\mathbb{Z}}^r=\mathbb{Z}{e}_1\oplus H^{\mathrm{\prime }}$. Every $p^{\mathrm{\prime }}\in P$ has a unique form $a{e}_1+h^{\mathrm{\prime }}$ with $h^{\mathrm{\prime }}\in H^{\mathrm{\prime }}$, and $h^{\mathrm{\prime }}=p^{\mathrm{\prime }}-a{e}_1\in P\cap H^{\mathrm{\prime }}$. Therefore $P=\mathbb{Z}{e}_1\oplus (P\cap H^{\mathrm{\prime }})$. The subgroup $P\cap H^{\mathrm{\prime }}$ is saturated in $H^{\mathrm{\prime }}$; its rank is $\mathrm{rank}P-1$, since the projection of $P$ to $\mathbb{Z}{e}_1$ is onto. Induction makes $P\cap H^{\mathrm{\prime }}$ a direct summand of $H^{\mathrm{\prime }}$, hence $P$ a direct summand of ${\mathbb{Z}}^r$.

Apply step 3.1 to the saturated rank-$g$ subgroup $M\subset H\cong {\mathbb{Z}}^{2g}$ from step 2.1. It is a direct summand of $H$.

Since $M\subset L$ and $V$ vanishes on $L\times L$, its restriction to $M\times M$ vanishes. Together with rank $g$ and the direct-summand conclusion of step 4.1, this proves the claim.