I am posting here the answer that I gave to the same question when it was posted yesterday on MSE.

Let $f_1$ and $f_2$ be independent functions on a symplectic manifold $(M,\omega).$

Let us denote by $\Sigma$ the submanifold $f_1^{-1}(0)\cap f_2^{-1}(0)$ of codimension $2$ in $M$.

The tangent bundle of $\Sigma$ is
$$T \Sigma= (\ker df_1\cap \ker df_2) |_\Sigma=(\text{span}\{X_{f_1},X_{f_2}\})^\perp|_\Sigma.\tag{1}$$

So in the symplectic vector bundle $(T_{\Sigma} M,\omega |_\Sigma)$ the vector sub-bundle $T\Sigma$ has orthogonal complement $$(T\Sigma)^\perp=\operatorname{span}\{X_{f_1},X_{f_2}\}|_\Sigma.\tag{2}$$

By definition, $\Sigma$ is symplectic in $(M,\omega)$ if and only $T\Sigma\cap(T\Sigma)^\perp=0 (\leftarrow\text{the zero section of }\Sigma).$

By (1) and (2), this means :
in any point of $\Sigma$ the linear system $$\left\{\begin{array}{c}0=\langle df_1,c_1X_{f_1}+c_2X_{f_2}\rangle=\{f_1,f_2\}c_2, \\0=\langle df_2,c_1X_{f_1}+c_2X_{f_2}\rangle=-\{f_1,f_2\}c_1\end{array}\right.$$ has only the trivial solution $c_1=c_2=0.$

Therefore $\Sigma$ is symplectic iff $\{f_1,f_2\}$ has no zeroes on $\Sigma.$