On the generalized Gauss-Bonnet theorem

When $E\to M$ is an oriented vector bundle of rank $2n$ over a compact manifold $M$, it has a well-defined de Rham Euler class $e(E)$ in $H^{2n}_{dR}(M)$, and a representative $2n$-form for $e(E)$ can be computed as follows:

Fix a positive definite inner product $\langle,\rangle$ on $E$. (Since any two such inner products are equivalent under automorphisms of $E$, it doesn't matter which one.) Let $\nabla$ be a $\langle,\rangle$-orthogonal connection on $E$, and let $K^\nabla$ denote the curvature of $\nabla$, regarded as a $2$-form with values in ${\frak{so}}(E)$. Let $e(\nabla) = \mathrm{Pf}(K^\nabla/2\pi)$, which is a well-defined $2n$-form on $M$. (N.B.: The definition of $\mathrm{Pf}$ requires both the inner product and the orientation of $E$.) Then $e(E)=\bigl[e(\nabla)\bigr]\in H^{2n}_{dR}(M)$. (That $\bigl[e(\nabla)\bigr]$ is independent of the choice of $\nabla$ is one of the first results proved in Chern-Weil theory.)

If $M$ is a compact, oriented $2n$-manifold, then the value of $e(E)$ on $[M]$, the fundamental class of $M$, can be computed as follows: Let $Y$ be a section of $E$ that has only a finite number of zeroes. (By Whitney transversality, the generic section of $E$ satisfies this condition.) Using the orientations of $E$ and $M$, one defines the index of an isolated zero $z$ of $Y$, which is an integer $\iota_Y(z)$. Then $$ e(E)\bigl([M]\bigr) = \int_M e(\nabla) = \sum_{z\in Z}\ \iota_Y(z). $$ By the Poincaré-Hopf Theorem, when $E = TM$ (as oriented bundles), this sum is equal to the Euler characteristic of $M$, which explains why $e(E)$ is called the Euler class of $E$.

To prove this result, one chooses an orthogonal connection $\nabla$ on $E$ that depends on $Y$ and whose Euler form $e(\nabla)$ is easy to evaluate explicitly. This can be done as follows:

Let $Z\subset M$ be the (finite) zero set of $Y$ and let $U\subset M$ be an open neighborhood of $Z$ that consists of disjoint, smoothly embedded $2n$-balls, one around each element of $Z$. Let $\phi$ be a smooth function on $M$ that is identically equal to $1$ on an open neighborhood of $Z$ and whose support $K\subset U$ is a disjoint union of closed, smoothly embedded balls, one around each element of $Z$.

Now, $E$ is trivial over $U$, so choose a positively oriented $\langle,\rangle$-orthonormal basis of sections $s_1,\ldots, s_{2n}$ of $E$ over $U$ and let $\nabla_1$ be the (flat) connection on $E_U$ for which these sections are parallel. Let $\bar Y = Y/|Y|$ be the normalized unit section defined on $M\setminus Z$, and define a second connection on $U$ by $$ \nabla_2 s = \nabla_1 s + (1{-}\phi)\bigl( \bar Y\otimes \langle s,\nabla_1 \bar Y\ \rangle - \langle s,\bar Y\ \rangle\ \nabla_1\bar Y\ \bigr). $$ It is easily verified that this formula does define a connection on $U$, that $\nabla_2$ is $\langle,\rangle$-orthogonal, and that, outside of $K$ (i.e., where $\phi\equiv0$), the vector field $\bar Y$ is $\nabla_2$-parallel. (Note that, because $\phi\equiv1$ near $Z$, where $Y$ is not defined, this formula does extend smoothly across $Z$, agreeing with $\nabla_1$ on a neighborhood of $Z$.)

Meanwhile, on $M\setminus K$, write $E$ as a $\langle,\rangle$-orthogonal direct sum $E = \mathbb{R}\cdot Y \oplus E'$ and choose a $\langle,\rangle$-compatible connection $\nabla_3$ on $E$ over $M\setminus K$ that preserves this splitting. Using a partition of unity subordinate to the open cover of $M$ defined by $U$ and $M\setminus K$, construct a $\langle,\rangle$-orthogonal connection $\nabla$ that agrees with $\nabla_2$ on $K$ and with $\nabla_3$ on $M\setminus U$, and for which the line bundle $\mathbb{R}\cdot Y\subset E$ is parallel on $M\setminus K$.

Now, on $M\setminus K$, the curvature of $\nabla$ takes values in ${\frak{so}}(E')\subset{\frak{so}}(E)$, so $e(\nabla) = \textrm{Pf}\bigl(K^{\nabla}/(2\pi)\bigr)$ vanishes identically outside of $K$. It suffices now to evaluate the integral of $e(\nabla)$ over a single component $B$ of $K$, which may be assumed to be the unit ball in $\mathbb{R}^{2n}$, so restrict attention to this case. Write $\bar Y = s_1 u_1 +\cdots + s_{2n}\ u_{2n}$, and note that, on $B$, one has, by definition, $$ \nabla s_i = \nabla_2 s_i = \sum_{j=1}^{2n} s_j\otimes (1{-}\phi)(u_j\ du_i - u_i\ du_j), $$ i.e., the connection $1$-forms of $\nabla$ in this basis are $\omega_{ij} = (1{-}\phi)(u_i\ du_j - u_j\ du_i)$.
Using the identity ${u_1}^2 +\cdots + {u_{2n}}^2 = 1$, the curvature forms are easily computed to satisfy $$ \Omega_{ij} = d\omega_{ij} + \omega_{ik}\wedge\omega_{kj} = (1{-}\phi^2)\ du_i\wedge du_j - d\phi\wedge(u_i\ du_j - u_j\ du_i). $$ At this point, you have to know the definition of the Pfaffian. (I'll wait while you look it up.) Using the fact that the result has to be invariant under $SO(2n)$-rotations, it is easy to show that, on $B$, one has $$ e(\nabla) = \textrm{Pf}\left(\frac{\Omega}{2\pi}\right) = c_n (1{-}\phi^2)^{n-1} d\phi\wedge u^*(\Upsilon) $$ for some universal constant $c_n$ and where $\Upsilon$ is the $SO(2n)$-invariant $2n$-form of unit volume on $S^{2n-1}\subset\mathbb{R}^{2n}$ and $u: B\setminus z\to S^{2n-1}$ is $u = (u_1,\ldots, u_{2n})$. (I'll let you evaluate the constant $c_n$. This can be done in a number of ways, but it's essentially a combinatorial exercise.) In particular, $$ e(\nabla) = d\bigl(P_n(1{-}\phi)\ u^*(\Upsilon)\bigr) $$ where $P_n(t)$ is the polynomial in $t$ (of degree $2n{-}1)$) that vanishes at $t=0$ and satisfies $P'_n(1{-}t) = -c_n(1{-}t^2)^{n-1}$. By Stokes' Theorem, one has $$ \int_B e(\nabla) = P_n(1) \int_{\partial B}u^*(\Upsilon) = P_n(1)\ \textrm{deg}(u:\partial B\to S^{2n-1}) = P_n(1)\ \iota_Y(z) $$ Thus, it follows that there is a universal constant $C_n = P_n(1)$ such that $$ e(E)\bigl([M]\bigr) = \int_M e(\nabla) = C_n \ \sum_{z\in Z}\ \iota_Y(z). $$ Now, $C_1 = 1$, as one can show by computing with the constant curvature metric on the $2$-sphere and noting that, for $Y$ the gradient of the height function on $S^2$, the sum of the indices is $2$. Finally, by properties of the Pfaffian and the index, one sees that, if such a formula as above is to hold, then one must have $C_{n+m}=C_nC_m$. Thus, $C_n=1$ for all $n$, and the formula is proved.

(Of course, one can avoid these final tricks for evaluating the constants by carefully doing the combinatorial exercise, but I'll leave that to the curious.)

You can try Chap. 13, vol. 5 of M. Spivak's opus A comprehensive Introduction to Differential Geometry. (On the cover of this volume there are three birds carrying a banner that reads "All the way with Gauss Bonnet")

Also you can try my book Lectures on the geometry of manifolds where I discuss many approaches to this theorem and connections to other problems in geometry.

For a really gentle introduction to this theorem I would also recommend the small survey paper The many faces of Gauss-Bonnet which is a talk that I gave to first year graduate students a few years ago. It contains many figures and plenty of other references.

Maybe what follows is not exactly what you are looking for, but it gives you an answer at least when you don't want to restrict yourself to the tangent bundle and work rather with general (complex) vector bundles.

All that I write you can find on "Principle of algebraic geometry", by Griffiths and Harris.

Let $M$ be a compact, oriented manifold, $E\to M$ a complex vector bundle of rank $k$ and $\sigma=(\sigma_1,\dots,\sigma_k)$ $k$ global smooth sections of $E$. Define the degeneracy set $D_i(\sigma)$ to be the set of points $x\in M$ where $\sigma_1,\dots\sigma_i$ are linearly dependent, that is $$ D_i(\sigma)=\{x\mid\sigma_1(x)\wedge\cdots\wedge\sigma_i(x)=0\}. $$ One says that the collection $\sigma$ is generic if, for each $i$, $\sigma_{i+1}$ intersects the subspace of $E$ spanned by $\sigma_1,\dots,\sigma_i$ transversely and if moreover the integration over $D_{i+1}(\sigma)\setminus D_i(\sigma)$ defines a closed current.

This happens, for instance, if everything is complex analytic and the dimensions are the expected ones.

Now, suppose that $\sigma_1,\dots,\sigma_k$ are generic sections of $E$. Then, the Gauss-Bonnet Formula reads:

The $r$th Chern class $c_r(E)$ is Poincaré dual to the degeneracy cycle $D_{k-r+1}$.

In particular, suppose that $M$ is of even dimension $2k$. Then, the top Chern class $c_k(E)\in H^{2k}(M,\mathbb Z)\simeq\mathbb Z$. If $\sigma$ is a smooth section of $E$ having non-degenerate zeros, then the integer $c_k(E)$ counts precisely these zeros (with sign, depending on orientations), which compose the degeneracy cycle $D_1$.

Finally, to recover the top Chern class of $E$ from its differential-geometric data, just recall that the Chern forms $c_r(E,\nabla)$ of a vector bundle $E$ endowed with a connection $\nabla$ are defined by the formula $$ \det(I+t\Theta(E,\nabla))=1+tc_1(E,\nabla)+\cdots+t^kc_k(E,\nabla), $$ where $\Theta(E,\nabla)$ is the curvature of the connection $\nabla$.

Then your integral quantity is given by $$ \int_M c_k(E,\nabla). $$

Specializing further, if $M$ is a compact complex manifold of dimension $k$ and $E=T_M$ its holomorphic tangent bundle, then $$ c_k(T_M)=c_k(M)=\chi_{\text{top}}(M). $$