Let $\nu: M \to \mathbb S^n$ be a normal unit vector field along $M$, then the derivative $d\nu$ of $\nu$ maps $T_pM$ to $T_{\nu(p)}S = \nu(p)^\perp = T_pM$ and the Gaussian curvature is given by $$K_p = \det(d\nu(p): T_pM \to T_pM)$$
Now the volume form on $M$ is given by $\mathrm{d}vol_M = \iota(\nu) \mathrm{d}vol_{\mathbb R^{n+1}}$, i.e. for tangent vectors $\xi_1,\dots, \xi_n \in T_pM$ we have $$\mathrm{d}vol_M(p)(\xi_1, \dots, \xi_n) = \mathrm{d}vol_{\mathbb R^{n+1}}(p)(\nu(p), \xi_1, \dots, \xi_n) = \det(\nu(p), \xi_1, \dots, \xi_n)$$
Now consider the pullback $\nu^\ast \mathrm{d}vol_{S^n}$ of $\mathrm{d}vol_{S^n}$ to $M$:
\begin{eqnarray*}
\nu^\ast \mathrm{d}vol_{S^n}(p)(\xi_1, \dots, \xi_n) &=& \mathrm{d}vol_{S^n}\left(\nu(p)\right)\left(d\nu(p) \xi_1, \dots, d\nu(p)\xi_n\right) \\ &=& \det\left(\nu(p), d\nu(p) \xi_1, \dots, d\nu(p)\xi_n\right) \\ &=& K_p\cdot \det\left(\nu(p), \xi_1, \dots, \xi_n\right) \\
&=& K_p \; \mathrm{d}vol_M
\end{eqnarray*}
Therefore $$\int_M K \; \mathrm{d}vol_M = \int_M \nu^\ast \mathrm{d}vol_{S^n} = \deg(\nu) \int_{S^n} \, \mathrm{d}vol_{S^n} = \deg(\nu) \cdot \text{Volume }S^n$$
For even $n$, we have $\deg(\nu) = \frac{\chi(M)}2$, so I guess one might consider this to be a generalization to odd dimensional manifolds.
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.
Best Answer
Every Riemannian manifold $(X,g)$ can be isometrically embedded into $\mathbb{R}^N$ for $N$ sufficiently large. So in principle, when one wants to prove something about Riemannian manifolds, it suffices to consider isometrically embedded submanifolds $X\subset\mathbb{R}^N$. However, this is not the most natural way to study the geometric object $(X,g)$, because it requires a choice of embedding into $\mathbb{R}^N$. Hence, this is called "extrinsic". The text states that Carl B. Allendoerfer and André Weil proved the theorem in full generality for embedded submanifolds $X\subset\mathbb{R}^N$, and thus for every Riemannian manifold - but not in the most natural way, because it requires a choice of embedding.
There is also a way to prove the same theorem without making reference to any embedding into Euclidean space. This is called "intrinsic", and the text states the Chern was the first to prove the theorem in full generality without making reference to such an embedding. Doing this does not prove anything new, because every Riemannian manifold can already be viewed as a submanifold of some $\mathbb{R}^N$. This is why the text says "despite the fact that Allendoerfer and Weil were the first to prove it in complete generality in higher dimensions." - Chern wasn't the first to prove the theorem, but he gave a more natural proof, which is why the result now bears his name.