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.
No, you need the Gaussian curvature of $M$ with the induced metric. (This is only going to be the product of the principal curvatures when $M$ is embedded in a flat space. Consider, for example, a great $2$-spheres in $S^3$, whose second fundamental form is identically $0$.) $K_{\text{sec}}=1$ for all $2$-planes in $T_pS^3$.
EDIT: There are classical formulas for the (intrinsic) curvature of a Riemannian surface in terms of its first fundamental form. This is also very nicely calculated using orthonormal moving frames to get the connection $1$-form $\omega_{12}$ and the intrinsic Gaussian curvature $K$ is given by the equation $-K\,dA = d\omega_{12}$. Indeed, using this equation, it's not hard to give a proof of the Gauss-Bonnet theorem using Stokes's Theorem.
In general, if $M\subset N$ is a submanifold of the Riemannian manifold $N$, then the Gauss equations relate the curvature of $M$ to the second fundamental form and the curvature tensor of $N$. In particular, when $N=S^3$ has constant curvature $1$, the formula you want is
$$ K = k_1k_2 + 1.$$
Note that this checks with the basic examples we were discussing in comments.
Best Answer
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.