Can the Gauss-Bonnet theorem be proven from Stokes's theorem?

In a comment to this question, John Ma claims that the Gauss-Bonnet theorem can be proven from Stokes's theorem, but does not explain how.

For two dimensions, Stokes's theorem says that for any smooth 2-manifold (i.e. surface) $S$ and one-form $\omega$ defined on $S$,

$$\oint_{\partial S} \omega = \iint_S d\omega.$$

I could vaguely imagine coming up with some kind of one-form $\omega$ that depends on the metric, such that (a) along the boundary curve $\omega$ maps the boundary tangent vector to the geodesic curvature and (b) in the surface interior $\ast d\omega$ equals the Gaussian curvature. (In more concrete vector-field language, this corresponds to a vector field $\vec{\omega}$ defined over the surface such that (a) on the boundary curve $\vec{\omega} \cdot d\vec{l}$ equals the curve's geodesic curvature and (b) in the surface interior $(\vec{\nabla} \times \vec{\omega}) \cdot d\vec{S}$ equals the Gaussian curvature.) This would reproduce part of the Gauss-Bonnet formula, but how could you possibly get out the Euler characteristic term?

See Differential Forms and Applications (by Manfredo P. Do Carmo ) chapter 6 section 1. The proof of Gauss-Bonnet's Theorem presented by Do Carmo in his text is essentially the same as given by S.S. Chern and use the Stokes' theorem. The original proof of S.S. Chern is found here.