Linear Maps from $V$ to $V^*$ defined by a 2-form I came across this idea at the very beginning of a book and I don't quite seem to grasp it. It states given $\omega \in \bigwedge ^2 (V) $  you can define a linear map $ \omega^\#: V \to V^* $ by $\omega(v, \cdot)$ .
My first question is just a verification. For example, if $\dim(V) = 2$ then $\omega = a e^1 \wedge e^2$. If we take a vector $v = v_1 e_1 + v_2 e_2$ would $\omega (v, \cdot) = v_1 e^2 - v_2 e^1 $ ?
While this seems all well and good, it continues on and states that it wishes to show that $\omega = dx \wedge dy$ is non-degenerate. It does this by saying to considered 
$$\omega ^\#  \left ( \frac{\partial }{\partial x} \right )  = \iota \left ( \frac{\partial }{\partial x} \right )  (dx \wedge dy) = dy $$
First, I am not certain where the iota came from or if it was a typo. 
Assuming $\frac{\partial}{\partial x} = \frac{\partial}{\partial x} e_1$, then from my first question, it would follow that $\left ( \frac{\partial }{\partial x} \right )  (dx \wedge dy) = \iota \frac{\partial }{\partial x}dy$. Why is this simply $dy$ or is my approach simply wrong?
Thanks again for the help. 

The symbol $\iota $  is a common notation for the insertion of a vector fields. It maps $k$-forms to $k-1$-forms having the eating the vector, that is \begin{equation}
\iota_v(\omega)(u_1, \ldots, u_{k-1})=\omega(v,u_1,\ldots,u_{k-1})
\end{equation}
Different conventions are around and sometimes the vector is fed into the last argument. So in your case 
\begin{equation}
\iota_{\partial/\partial x}\ \omega=\omega(\partial/\partial x, \cdot)=(\mathrm{d}x\wedge\mathrm{d}y)(\partial/\partial x,\cdot)=\mathrm{d}y
\end{equation}
