Good question in symplectic linear algebra. $f$ must be linear (even if we do not assume it to be bijection) for $dim(V)\ge 4$.

First remark is that the restriction of $f$ to any isotropic subspace is linear.

V is symplectomorphic $L^*\oplus L$ with the symplectic structure $\omega ((a,b),(x,y))=a(y)-y(b)$. We may assume (after subtraction of linear map $f|_{L^*\oplus 0}+f|_{0\oplus L}$) that $f$ equals to zero on $L^*\oplus 0$ and on $0\oplus L$. It remains to prove that $f$ is totally zero map.

Note that $f$ is automatically zero on such $(a,b)$ that $a(b)=0$ (since (a,b)=(a,0)+(0,b)). Denote the set of all such vectors by $Ann$. Now I claim that any vector $(x,y)$ ($x(y) \ne 0$) can be decomposed as a sum of two skew-orthogonal vectors from $Ann$ and hence the value of $f$ is zero on $(x,y)$.

There are plenty of opportunities to construct such a decomposition. Obviously, it is sufficient to prove the existence of such a decomposition for $x(y)=1$. We can choose a basis $e_1,..$ in $L$ and dual basis $f^1,..$ in $L^*$ such that $w=(x,y)=(f^1,e_1)$. Now take decomposition $(f_1,e_1)=1/2(f^1+f^2,e_1-e_2)+1/2(f^1-f^2,e_1+e_2)$. It is easy to check that vectors $1/2(f^1+f^2,e_1-e_2)$ and $1/2(f^1-f^2,e_1+e_2)$ belong to $Ann$ and skew-orthogonal. So we are done.

Good question in symplectic linear algebra. $f$ must be linear (even if we do not assume it to be bijection) for $dim(V)\ge 4$.

First remark is that the restriction of $f$ to any isotropic subspace is linear.

V is symplectomorphic to $L^*\oplus L$ with the symplectic structure $\omega ((a,b),(x,y))=a(y)-x(b)$. We may assume (after subtraction of linear map $f|_{L^*\oplus 0}+f|_{0\oplus L}$) that $f$ equals to zero on $L^*\oplus 0$ and on $0\oplus L$. It remains to prove that $f$ is totally zero map.

Note that $f$ is automatically zero on such $(a,b)$ that $a(b)=0$ (since (a,b)=(a,0)+(0,b)). Denote the set of all such vectors by $Ann$. Now I claim that any vector $(x,y)$ ($x(y) \ne 0$) can be decomposed as a sum of two skew-orthogonal vectors from $Ann$ and hence the value of $f$ is zero on $(x,y)$.

There are plenty of opportunities to construct such a decomposition. Obviously, it is sufficient to prove the existence of such a decomposition for $x(y)=1$. We can choose a basis $e_1,..$ in $L$ and dual basis $f^1,..$ in $L^*$ such that $w=(x,y)=(f^1,e_1)$. Now take decomposition $(f_1,e_1)=1/2(f^1+f^2,e_1-e_2)+1/2(f^1-f^2,e_1+e_2)$. It is easy to check that vectors $1/2(f^1+f^2,e_1-e_2)$ and $1/2(f^1-f^2,e_1+e_2)$ belong to $Ann$ and skew-orthogonal. So we are done.