If $\omega$ vanishes in a surface, so do $d\omega$

Let $S$ be that surface and $\iota : S\to U$ be the inclusion. The fact that $\omega$ vanishes over the tangent vectors of $S$ is the same as saying that $\iota^*\omega = 0$.

With that in mind, the proof is trivial: Since pullback $\iota^*$ commutes with exterior differentiation $d$,

$$ \iota^* (d\omega) = d (\iota^* \omega) = d(0) = 0.$$

Another proof: Recall the well-known formula for exterior derivative $$ d\omega(X_1,X_2,X_3,\dots,X_{r+1})=\sum_i(-1)^{i+1} X_i\omega(X_1,\dots,\widehat{X_i},\dots,X_{r+1})+\sum_{i<j}(-1)^{i+j}\omega([X_i,X_j],X_1,\dots,\widehat{X_i},\dots,\widehat{X_j},\dots,X_{r+1}). $$ Now if $\omega$ vanishes on $M$, then for all $X_1,\dots,X_{r+1}\in\mathcal{X}(M)$, we have $[X_i,X_j]\in\mathcal{X}(M)$ and hence $\omega([X_i,X_j],X_1,\dots,\widehat{X_i},\dots,\widehat{X_j},\dots,X_{r+1})=0$ by supposition, $X_i\omega(X_1,\dots,\widehat{X_i},\dots,X_{r+1})$ is differentiating constant $0$ so is $0$, and so $d\omega$ vanishes on $M$.