About Gauss-Bonnet Theorem

Good question.

The answer is that Gauss-Bonnet does not actually require the hypothesis that $\Sigma$ is embedded (or even immersed) in $\mathbb{R}^3$. Rather, it is an intrinsic statement about abstract Riemannian 2-manifolds. See Robert Greene's notes here, or the Wikipedia page on Gauss-Bonnet, or perhaps John Lee's Riemannian Manifolds book.

Many texts on elementary differential geometry include the hypothesis that $\Sigma \subset \mathbb{R}^3$ because that is the context they're working in. That is, some books don't define abstract manifolds.

In this case, I would interpret $K$ as the sectional curvature. To be honest, I've never seen a definition of "principal curvature" outside the setting of $\mathbb{R}^N$, though my ignorance shouldn't be taken as definitive proof that the concept doesn't exist in more general settings.

Aside: More interesting to me, by the way, is the fact that $\Sigma$ is non-orientable (it is homeomorphic to $\mathbb{RP}^2$). I don't think I've seen a proof of Gauss-Bonnet for the non-orientable setting, but it's apparently not a difficult modification.

I think a proof of Gauss-Bonnet for non-orientable setting is very simple. Here we follow John Lee's idea in his book (page 169).

Let $M$ be a compact two dimensional non-orientable manifold. Then a double cover of $M$ must a compact, orientable two dimensional manifold. The double cover $\tilde{M}$ should 'lift up' the piece corresponding to the $\mathbb{RP}^{2}$ or $K$ in the classification theorem. Now for $\tilde{M}$ we can endow the push-back metric defined by $$ \tilde{g}=\pi^{*}g $$ such that the projection map is a local isometry. Then it is obvious that $\pi^{*}\tilde{K}=K$ since sectional curvature is only defined locally. Now we have $$ \chi(\tilde{M})=\frac{1}{2\pi}\int_{\tilde{M}} \tilde{K}d\tilde{\sigma} $$ But we know that that $\chi(\tilde{M})=2\chi(M)$, and similarly $\tilde{K}=K$'s integral would appear twice when we do the integral on $M$ using a compact-supported density. So dividing by 2 on both sides we get the desired formula for $M$: $$ \chi(M)=\frac{1}{2\pi}\int_{M} Kd\sigma $$ and this finished the proof.