3 added 292 characters in body

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 exactly the number a smooth section of $E$ having non-degenerate zeros, then the integer $c_k(E)$ counts precisely these zeros (counted with multiplicities) of a generic section of sign, depending on orientations), which compose the degeneracy cycle $E$.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 $$c_k(T_M)=c_k(M)=\chi_{\text{top}}(M).$$ 2 added 217 characters in body; added 75 characters in body 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$c_k(E)\in H^{2k}(M,\mathbb Z)\simeq\mathbb Z$, is exactly the number of zeros (counted with multiplicities) of a generic section of$E$. 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 $$\int_M c_k(E,\nabla).$$ 1 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$c_k(E)\in H^{2k}(M,\mathbb Z)\simeq\mathbb Z$, is exactly the number of zeros (counted with multiplicities) of a generic section of$E$. 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\$.