Let $\phi$ be a bilinear form on the vector space that has basis $\{x_i\}\cup\{y_j\}$.
If we have $x\in \operatorname{Span}\{x_i\}$ and $y\in\operatorname{Span}\{y_i\}$ are such that $\phi\left(x\,\,\,,\,\,\,\operatorname{Span}\{x_i\}\right)+\phi\left(y\,\,\,,\,\,\,\operatorname{Span}\{y_i\}\right)=0$,
[Edit: As suggested by @awllower, I should indeed explain my notation: By taking $\operatorname{Span}\{v_i\}$, $v=x, y$ as a variable of the bilinear form, I mean that for all $v\in \operatorname{Span}\{v_i\}$ the relation is true.]
Given that $\phi$ is non-degenerate on $\operatorname{Span}\{x_i\}$
Can I conclude that $x=0=y$?
No, you can't. You could have $\phi(y,z)=0$ for all $z$ despite $y\ne0$, and then your condition would be fulfilled with $x=0$ and $y\ne0$.