$i^*(\omega)$ vanish on S $\Leftrightarrow$ A is tangent to $S$ If $\omega\in \Omega^2(\mathbb{R}^3)$ is a 2-form: $\omega=f\;dx\wedge dy+g\;dx\wedge dz+h\;dy \wedge dz$, and $S$ is a surface in $\mathbb{R}^3$, prove that $i^*(\omega)$ vanish on $S$ $\iff$ $A=h\frac{\partial}{\partial x}-g\frac{\partial}{\partial y}+f\frac{\partial}{\partial z}$ is tangent to $S$.
The only clue I have been able to find is that there exists this isomorphism:
$$ \phi:\mathfrak{X}(\mathbb{R}^3)\longrightarrow \Omega^2(\mathbb{R}^3)$$
$$ A \longmapsto \phi(A),$$
where $\phi(A)(B,C)=(B\times C)\cdot A$. In this way $\phi(A)=h\;dx\wedge dy-g\;dx\wedge dz+f\;dy \wedge dz$, which is similar to what we are working about. Any idea?

By duality, $\omega$ corresponds to the vector field $(h,-g,f)$. Then $\iota^\ast(\omega)$ measures exactly the normal component of $(h,-g,f)$ with respect to $S$. With this in mind, there's nothing else to do.
Explicitly, we identify $$F_1\partial_x + F_2\partial_y +F_3 \partial_z \leftrightarrow F_1 \;{\rm d}y\wedge dz +F_2 \,{\rm d}x\;\wedge {\rm d}y + F_3\; {\rm d}x \wedge {\rm d}y.$$You can check that this is an isomorphism between $\mathfrak{X}(\Bbb R^3)$ and $\Omega^2(\Bbb R^3)$. This works for the following reason: if $\beta_2$ denotes that isomorphism, and $$\begin{align} {\rm id}\colon & \quad C^\infty(\Bbb R^3) \ni f \mapsto f \in \Omega^0(\Bbb R^3)\\ \beta_1:& \quad \mathfrak{X}(\Bbb R^3) \ni F_1\partial_x+F_2\partial_y+F_3\partial_z \mapsto F_1\,{\rm d}x+F_2\,{\rm d}y+F_3\,{\rm d}z \in \Omega^1(\Bbb R^3) \\ \beta_3:&\quad C^\infty(\Bbb R^3) \ni f \mapsto f\,{\rm d}x\wedge {\rm d}y\wedge {\rm d}z \in \Omega^3(\Bbb R^3),\end{align}$$then $\beta_1 \circ {\rm grad} = {\rm d} \circ {\rm id}$, $\beta_3\circ {\rm div} = {\rm d} \circ \beta_2$ and $\beta_2 \circ {\rm curl} = {\rm d} \circ \beta_1$. Also, if $F,G \in \mathfrak{X}(\Bbb R^3)$, we have $$\beta_1(F)\wedge \beta_1(G) = \beta_2(F \times G).$$
