Given a finite $p$-group $A$, with a non-degenerate quadratic form $q:A\rightarrow \mathbb Q/2\mathbb Z$ (that is a map satisfying $q(na)=n^2q(a)$ for all $n\in \mathbb Z,a\in A$), an important result of Nikulin (see his paper "Integral symmetric bilinear forms and some of their applications") is that there exists a unique $p$-adic lattice $K(q_p)$ of rank $l(A)$ (that is the minimal number of generators of $A$) whose discriminant-form is isomorphic to $q$ except in two exceptional cases when $p=2$. Here the discriminant form is the quadratic form obtained from extending $q_p$, the quadratic form on $K(q_p)$, to $K(q_p)^*$ and then restricting to $K(q_p)^*/K(q_p)$.

My question is, when $p=2$, how can you determine what $K(q_2)$ is? It would seem that we need more information than just the values taken by the discriminant form. For example, consider the orthogonal sum $q_1^{(2)}(2)\oplus q_1^{(2)}(2)$, where $q_1^{(2)}(2)$ is the quadratic form on the rank 1 $2$-adic lattice given by $q(v)=2v^2$. The values taken by the discriminant form are easily seen to be $0,1,1,1$ on $\mathbb Z/2\oplus \mathbb Z/2$. Similarly, consider the rank 2 $2$-adic lattice with quadratic form $q(x,y)=4xy$. The discriminant group is again $\mathbb Z/2\oplus \mathbb Z/2$, and the discriminant form takes the same values $0,1,1,1$. According to the Proposition 1.11.2 in that paper, these two lattices have different signs, so how can one differentiate between them?

That is, given just the discriminant group and quadratic form, how can we determine which $2$-adic lattice we're dealing with?