Let $V,W \leq X'$, where $X$ is a vector space and $X'$ its dual. Let $f \in X'$. How do I check if $f \in V \wedge W$?
To make it concrete, Let $X$ be a real vector space with complex structure $J$, $B: X \times X \rightarrow \mathbb{R}$ a symmetric bilinear form with $B(J \cdot, J \cdot)=B(\cdot,\cdot)$. Why is then the alternating bilinear form on the complexification $w(\cdot,\cdot)=B_\mathbb{C}(J\cdot,\cdot)$ in $V\wedge W$, where $V$ vanishes on the $+i$-eigenspace of $J$, and $W$ on the $-i$-eigenspace.
Suppose that $V=Y_1^{\perp}$ and $W=Y_2^{\perp}$, where $Y = Y_1 \oplus Y_2$ is a vector space. Then $$V \wedge W = \lbrace \phi \in Y' \wedge Y' : \phi(Y_1 , Y_1) = \phi(Y_2 , Y_2)=0 \rbrace :=S$$ (The definition of $S$ uses the standard translation between elements of $Y' \wedge Y'$ and alternating bilinear forms on $Y$).
Proof: It is clear that $V \wedge W \subset S$. Conversely, let $\lbrace a_i \rbrace$ and $\lbrace b_j\rbrace$ be bases for $Y_1$, $Y_2$ respectively, and let $\lbrace \alpha_i, \beta_j \rbrace$ be the basis $B$ of $Y'$ dual to the basis $\lbrace a_i, b_j \rbrace$ of $Y$. Then $\lbrace \beta_j \rbrace$ and $\lbrace \alpha_i \rbrace$ are bases for $V$ and $W$, respectively.
To show that $\phi \in S$ is contained in $V \wedge W$, it suffices to show that when $\phi$ is written in terms of the basis of $Y' \wedge Y'$ determined by $B$, all of the coefficients of the basis vectors of the form $\alpha_i \wedge \alpha_{i'}$ or $\beta_{j} \wedge \beta_{j'}$ are zero. This is clear from the definition of $S$.
To apply this to your case, let $(X,J)$ be a finite-dimensional real vector space with complex structure, let $Y=X_{\mathbb{C}}$, and let $Y_1$ and $Y_2$ be the the $i=\sqrt{-1}$ and $-i$ eigenspaces of $J_{\mathbb{C}}$ inside $Y$, respectively. One knows that $w\in S$, since for $v_1, v_2 \in Y_1$, $$w(v_1,v_2)=w(Jv_1, Jv_2) = w(iv_1,iv_2)=i^2w(v_1,v_2)=-w(v_1,v_2)$$ implying that $w(v_1,v_2)=0$. Similarly, $w(Y_2,Y_2)=0$. Hence $w \in S = V \wedge W$.