Consider a finite dimensional vector space $V = F^n$ for some field $F$. (For simplicity you can assume $F= \mathbb R$ or $F= \mathbb C$.) Now consider a bilinear form
$$b : \begin{array}{l} \\ V \times V \to F \\ (u,v) \mapsto b(u,v) = u^t A v\end{array}$$
with $A \in F^{n \times n}$.
Now let us identify $V \times V = F^{2n}$. We now construct a "shuffling function" $\varphi : V \times V \to V \times V$ that permutes the $2n$ scalar parameters. For example for $n=2$ we could have $\phi(x_1,x_2,x_3,x_4) = (x_2,x_1,x_4,x_3)$. Finally we define the function
$$f = b \circ \varphi : V \times V = F^{2n} \to F$$
So we consider $f$ as a function of $2n$ arguments in $F$.
Just by being able to evaluate $f$ at arbitrary points, is it possible to determine which $n$ of the $2n$ arguments go into the one or the other (left "$u$" or right "v") argument of $b$?
The answer to that in general is no, which is easy to see for the cases $A=0$ or $A=I$.
But: Can we find all matrices $A$ for which the question above can be answered with yes? (Or can we at least find some class of matrices $A$ for which this holds?) I'd be also happy for answers that just cover the case of a specific field like $F= \mathbb R, \mathbb Q, \mathbb C$.
This is just one class of matrices, it is no complete classification: The class of matrices without any zero entries.
Consider the exapmle for $n=2$ in $F = \mathbb R$: Let's say we evaluate following values which let us "probe" the matrix entries:
$$\begin{align}f(1,0,1,0) &= 0 \\ f(1,1,0,0) &= 1 \\ f(1,0,0,1) &= -1 \\ f(0,1,1,0) &= 1 \\ f(0,0,1,1) &= -1 \\ f(0,1,0,1) &= 0\end{align}$$
It is immediately clear that $A$ is a matrix without any zeros and the only way we do get zeros is if (at least) one side (e.g. $u$) is a zero vector. In this case we know that the first and third argument form one side, while the second and fourth argument form together the other side.
This argument can be applied to any matrix without zero entries of any size, and independent of the field.
(The example above was generated by $A= \begin{bmatrix} 1 & -1 \\ 1 & -1 \end{bmatrix}$ and $\varphi(x_1,x_2,x_3,x_4) = (x_1, x_3, x_2, x_4)$.)