So your $S$ is a subspace of the space of all skew-symmetric bilinear forms? I still don't understand the question, since a subspace of $V$ is isotropic under all elements of $S$ if and only if it is isotropic under all $\beta_i$, so the conclusion seems trivial from the assumption than the set of $\beta_i$ has no isotropic subspace of dimension $2$.

What do you mean by "the $\mathbb{Q}$-space generated by $\{\beta_i\}$"?

Is there a reason you use $n$ both for the number of colors and the width of the matrix? Can these be different numbers?

Just to be sure, you don't assume that the columns of $V_i$ are orthonormal? Because if you just assume orthogonality and not normality $V_iV_i^T$ is not an orthogonal projection.

You say that you want to find bounds in terms of $n$. But if you fix $n$ in my example you fix the lattice, so clearly the number of intersection points is finite. I'm not sure what you want your bounds to depend on in the general case.

What is $n$ for a general lattice?

There's a formula for the number of integer points on the circle of radius $\sqrt{n}$, and it is unbounded. Hence for suitable $n$ the lattice $\frac{1}{\sqrt{n}}\mathbb{Z}^2$ can intersect the unit circle in arbitrarily many points. Did I understood your question correctly?

Well, in the case of the sphere there are lattices intersecting it in an arbitrarily large number of points.

The condition for $a_{ij}=1$ is not very clear. Does it mean "there exists $1\le l \le t$ such that $v_i(l) \cap v_j(l) = \emptyset$", or "the union over $1\le l \le t$ of $v_i(l) \cap v_j(l)$ is empty" (which is the same as $v_i(l) \cap v_j(l) = \emptyset$ for all $l$), or something else?

If $B$ is the matrix of a bilinear form then this is just a change of basis.

Would you mind expanding on the combinatorial interpretation of $a_k$ and $b_k$?

Wow, that's a nice generalization, and your elementary proof is pretty neat (it's long because you wrote absolutely all the small details, but the big idea is quite simple).