This will not always work. Let $\{ e_1, e_2\}$ be the standard basis of $\mathbb R^2$ and rotate by 45 degrees, say, to obtain the basis $\{ f_1, f_2 \}$. Let $X$ be the smallest set that contains $(e_1,e_1)$, $(e_2,e_2)$, $(f_1,f_2)$, $(f_2,f_1)$ and satisfies your conditions (so we can put arbitrary constants in front of these vectors, but that's it since the third condition never applies). Then for example $(f_1,f_1)\notin X$.
However, it's easy to check that these four vectors are a basis of the tensor product (the first two and the last two are linearly independent, and the corresponding subspaces don't intersect), so every bilinear form that vanishes on $X$ is identically equal to zero.