If the function $\gamma\colon V\times V\to V$ satisfies
for all $u,v,x,y\in V$ and all scalars $a$, then $\gamma(u,v)=0$ for all $u,v$.
Indeed, $\gamma(u,0)=\gamma(u,0\cdot0)=0\gamma(u,0)=0$ and similarly $\gamma(0,v)=0$; therefore
Note that this is basically the statement you proved. However, for a bilinear form property $(1)$ is not required. Rather, it is required that
which is very different from the other property.
Are there maps $f\colon V\times V\to V$ that satisfy property $(1)$? Yes, for instance all linear maps $V\oplus V\to V$ do. However, they don't satisfy $(2)$ and $(3)$, but only