The exterior derivative commutes with restriction

Fix $p\in U$ and tangent vectors $v_0,\dots ,v_k\in T_{p}M=T_{p}U$. Choose a chart around $p$ contained in $U$, extend each vector using constant coordinate coefficients there, multiply by a bump equal to one near $p$, and extend by zero. This gives smooth fields $Y_0,\dots ,Y_k$ on $M$ with $Y_i(p)=v_i$.

Evaluate the invariant formula for $d_M\omega$ on $Y_i$ and the formula for $d_U(\omega {\mathrm{\mid }}_U)$ on $Y_i{\mathrm{\mid }}_U$. Each scalar evaluation of the form restricts to the same smooth function on $U$, so its directional derivatives agree there. Brackets restrict as well: their commutators on smooth functions have identical local expressions. Thus every term in the two formulas has the same value at $p$.

The arbitrary tangent vectors in step 1.1 show equality of the two $(k+1)$-covectors at $p$, and arbitrary $p$ gives $d_U(\omega {\mathrm{\mid }}_U)=(d_M\omega ){\mathrm{\mid }}_U$. Pullback by the open inclusion is restriction because its tangent map is the identity under $T_{p}U=T_{p}M$, proving the claim. The assertion is vacuous if $U$ is empty.