I am assuming you write perpendicular in body of the question to mean transversal as in its title. Now I read your question as searching for local functions $\hat{x}_2=\hat{x}_2(x_2,y_2),\hat{y}_2=\hat{y}_2(x_2,y_2)$ such that $dx_2\wedge dy_2=d\hat{x}_2\wedge d\hat{y}_2$ and $\mathcal{M}=\hat{x}_2^{-1}(0)\cap\hat{y}_2^{-1}(0)$ locally around $0$.
If this reading is correct then in general the answer is no.
Infact if there were local symplectic coordinates $(\hat{x}_2,\hat{y}_2)$ on the $(x_2,y_2)$ plane such that $\mathcal{M}\subseteq\hat{x}^{-1}_2(0)\cap\hat{y}^{-1}_2(0)$ in a neighborhood of $0$, then $\mathcal{M}$ should be locally included in the $(x_1,y_1)$ plane (because $\hat{x}^{-1}_2(0)\cap\hat{y}^{-1}_2(0)=x^{-1}_2(0)\cap y^{-1}_2(0)$.)
But one can construct a counter-example already taking symplectic vector subspaces complementary to the $(x_2,y_2)$-plane in the constant symplectic space $(\mathbb{R}^4,dx_1\wedge dy_1+dx_2\wedge dy_2)$.