Consider the 2-form $$\omega=\sum_{i=1}^ndx^i\wedge dy^i \,$$ on $\mathbb{R}^{2n}$, with coordinates $(x^1,\ldots,x^n,y^1,\ldots,y^n)$.
I want to be able to compute $\omega(\alpha,\beta)$, for $\alpha,\beta\in T_p\mathbb{R}^{2n}$.
I don't understand how to calculate, for example, $$\omega\left(\left(\frac{\partial}{\partial x_i}\right)\Big\vert_p,\left(\frac{\partial}{\partial y_j}\right)\Big\vert_p\right)\, (*)$$ I know for $1$-forms that $dx^i\frac{\partial}{\partial x_j}=\delta_{ij}$. I think that $(*)$ is also probably going to be $\delta_{ij}$, but I don't know how to show it.
The key thing here is the definition of $\wedge$: depending on convention, $$dx \wedge dy = dx \otimes dy - dy \otimes dx,$$ or it's half that. Going with this definition, we then have $$(dx^i \wedge dy^i)\left( \frac{\partial}{\partial x^j} , \frac{\partial}{\partial y^k} \right) = (dx^i \otimes dy^i)\left( \frac{\partial}{\partial x^j} , \frac{\partial}{\partial y^k} \right) - (dy^i \otimes dx^i)\left( \frac{\partial}{\partial x^j} , \frac{\partial}{\partial y^k} \right) \\ = dx^i\left( \frac{\partial}{\partial x^j} \right) dy^j\left( \frac{\partial}{\partial y^k} \right) - dy^i\left( \frac{\partial}{\partial x^j} \right) dx^j\left( \frac{\partial}{\partial y^k} \right).$$ The latter term is zero since the $x$ and $y$ are different coordinates. The duality implies that the first term is $\delta^i_j \delta^i_k$, and summing this over $j$ gives $\delta_{jk}$. One can do exactly the same thing for $\left( \frac{\partial}{\partial y^j} , \frac{\partial}{\partial x^k} \right)$, $\left( \frac{\partial}{\partial x^j} , \frac{\partial}{\partial x^k} \right)$ and $\left( \frac{\partial}{\partial y^j} , \frac{\partial}{\partial y^k} \right)$ (finding the former is $-\delta_{ik}$ and the latter two zero), and then use linearity on $\sum_{i} \alpha^i \frac{\partial}{\partial x^i} + \alpha'^i \frac{\partial}{\partial y^i}$ to arrive at $\omega(\alpha,\beta) = \sum_i (\alpha^i \beta'^i-\beta^i\alpha'^i)$.
Write $\alpha=(a_1,...,a_n,a'_1,...,,a'_n)$ and $\beta=(b_1,...,b_n,b'_1,...,b'_n)$ you have $\omega(\alpha,\beta)=\sum_i a_ib'_i-b_ia'_i$.