Here is an example:
Let $M=\mathbb{R}^3$ be a smooth manifold, and $S\subset M$ be an embedded submanifold with codimension $1$(say, $S$ is the unit sphere). I suppose $\exists V\in\mathfrak{X}(M)$ that is nowhere tangent to $S$ so that $S$ is orientable. Let $\omega=xdy\wedge dz + ydz\wedge dx + zdx\wedge dy \in\Omega^2(M) $ and $X = x\frac{\partial}{\partial x} + y\frac{\partial }{\partial y} + z\frac{\partial}{\partial z} \in \mathfrak{X}(M)$, then the integration of $\omega$ over $S$, which is $\int_S\iota^*\omega$ with $\iota$ being the inclusion map, will eqaul to the classical surface integration of $X$ over the surface $S$, if we choose proper orientation.
Can anyone give me a straightforward explanation from definitions why these two are equal?
This is because the classical surface integration of $X$ on the surface $S$ can be written as $$\int_S\vec{X}\cdot d\vec{S} = \int_S\iota^*[X\lrcorner(dx\wedge dy\wedge dz)] $$ where the right side stands for the integration on manifolds. Because $X\lrcorner(dx\wedge dy\wedge dz)=\omega$, we have $$\int_S\vec{X}\cdot d\vec{S}=\int_S \omega$$.