Here is a small comment to Seva's answer, which allows to get smaller bounds for some fields.
Assume that we managed to find functions $f,g$ so that $\sum f(a)W(a)=0$ if $\deg W\leqslant \alpha$, $\sum g(a) W(a)=0$ if $\deg W\leqslant \beta$. Then we have $$ \sum_{a\in A,b\in B} f(a)g(b)Q(a,b)=0 $$ for any polynomial $Q$ of degree at most $\alpha+\beta+1$. Indeed, if $Q$ is monomial, then our sum factorizes and one of factors is 0.
Apply this for a polynomial $Q(x,y)=P(x-y)$, where $\deg P\leqslant \alpha+\beta+1$ and $P(a,b)=0$ for $a,b\in A$, $a\ne b$. We get $$ 0=\sum_{a\in A,b\in B} f(a)g(b)P(a-b)=P(0)\cdot \sum f(a)g(a). $$ Now if $\sum f(a)g(a)\ne 0$, we conclude that $P(0)=0$. Assume for simplicity that $\alpha=\beta$. Then functions $f,g$ satisfying our condition are chosen in the same linear space of dimension at least $|A|-\binom{n+\alpha}\alpha$ (or $|A|-\sum_{i\leqslant \alpha} \binom{n}i$, if we consider only multilinear polynomials). If $P(0,0)\ne 0$, then this space is an isotropic subspace of $\mathbb{F}^{A}$ (any two vectors must be orthogonal.) Maximal dimension of isotropic subspace is well known subject and it always does not exceed $|A|/2$ be obvious reasons (such a space is contained in its own orthogonal complement). For real field there is no isotropic subspace even of dimension 1.
See the answer by Robin Chapman to my old question here for references on isotropic subspaces.