$i_\mathbb{X} \omega $ is the contraction of $\omega$ with respect to $\mathbb{X}$.
In my notes it is stated that $i_\hat{\mathbb{X}} dx = dx(\hat{\mathbb{X_t}})$. I cannot see how this fits the definition of contraction which fixes the first "coordinate" in a $k$-tuple.
Further apparently $dx(\hat{\mathbb{X_t}})$ is the first coordinate of $\hat{\mathbb{X_t}}$. I dont see how this could a we are taking an exterior derivative $d$.
So $dx$ is a $1$-form on $\Omega\subset \mathbb R^3$ for example where $(x,y,z)$ define the local coordinates. It means that $dx:\Omega\rightarrow \mathcal L(\mathbb R^3,\mathbb R)$ and $dx$ is simply the differential of the map $$\begin{array}{rccl}x:&\Omega&\rightarrow &\mathbb R \\ & p=(p_1,p_2,p_3)& \mapsto& p_1\end{array}$$
This differential is obviously given by $dx_p(u_1,u_2,u_3)=u_1$.
Now let $X=(X^1,X^2,X^3)$ be a vector field on $\Omega$, then $\iota_X(dx)$ is a $0$-form on $\Omega$ because the contraction reduces the degree by one of differential forms and using its definition, $$(\iota_X(dx))_p=dx_p(X(p))=X^1(p).$$