Let $V$ be a (finite dimensional) vector space over $K$, and $B: V \times V \to K$ a bilinear form. Consider a decomposition $V=V_1 \oplus V_2$, into vector subspaces $V_1,V_2$ of $V$.
In general, if we have bilinear forms $B_1, B_2$ on vector spaces $V_1,V_2$, then we have a bilinear form $\hat{B}: V_1 \oplus V_2 \times V_1 \oplus V_2 \to K$ defined by $\hat{B}((x_1,y_1),(x_2,y_2))=B_1(x_1,x_2)+B_2(y_1,y_2)$
Now clearly $B$ restricts to a bilinear form on each of $V_1,V_2$. Then it's natural to identify $\hat{B}$ (induced by $B|_{V_1}$ and $B|_{V_2}$) with $B$ on $V$ (right??)
But now let $x \in V_1, y\in V_2$ then $B(x,y)=\hat{B}((x,0),(0,y))=B(x,0)+B(0,y)=0$, which does not make sense as $B$ is just an arbitrary bilinear form on $V$.