I am trying to show that the euler class of the tangent bundle of a compact oriented connected manifold satisfies the following relation $\chi(TM)=(\sum_{i=1}^n \operatorname{ind}_{p_i}X)\mu$, where $X$ is a vector field with a finite number of zeros and $\mu \in H^d(M)$ is the canonical generator. The definition of the $\operatorname{ind}_p X$ is by considering a local representative of the vector field and the Gauss map of the representative $X: S_\varepsilon \rightarrow S^{d-1}$ that sends $x\rightarrow \frac{X(x)}{\|X(x)\|}$. The degree is defined in term of top degree cohomology. That is if we have two $n$-dimensional compact connected oriented manifolds $M,N$ and a map $\phi : M\rightarrow N$, then the degree of $\phi$ is the unique integer such that $\int_M \phi^* \omega=\deg(\phi)\int_N \omega$ for $\omega\in \Omega^d(N)$.
Basically let $\omega\in H^d_c(TM)$ be a form representing the Thom class. We want to check that $\int_M X^* \omega = \sum_i \operatorname{Ind}_{p_i}X$. Let's choose coordinate systems $(U_i,\phi_i)$ centered at $p_i$ and let's denote by $D_i:=\phi_i^{-1}(D^d)$. We can assume that $X_p\notin \operatorname{supp} \omega$ for $p\notin \bigcup_{i=1} D_i$. So we want to check that $\int_{D_i}X^*\omega = \operatorname{Ind}_{p_i}X$. Now to do this I am having some trouble. My idea would be the following $\int_{D_i}X^*\omega=\int_{D^d} (D\phi_i \circ X\circ \phi_i^{-1})^* \circ (D\phi_i^{-1})^* \omega$, and now use something along the lines of the Stokes's theorem to pass to an integral over $S^{d-1}$, but I am have a problem which is that the map that is doing the pull-back won't be exactly what I want since there is no division by the norm hapenning. Does this seem to be the right way to approach this? If so how do I confront this problem, if not how can one think about this in another way?
New edit : We could also use the fact that we are over a trivilizaing chart over the tangent bundle and there $H^d(D_i\times \mathbb{R}^n )=\{0\}$, and so $\omega=d\alpha$, and try and use this so that we get that $\int_{D^d} (D\phi_i \circ X\circ \phi_i^{-1})^* \circ (D\phi_i^{-1})^* \omega=\int_{S^{d-1}}(D\phi_i \circ X\circ \phi_i^{-1})^* \circ (D\phi_i^{-1})^* \alpha=(?)Ind_{p_i}X \int_{S^d} (D\phi_i^{-1})^* \alpha=Ind_{p_i}X\int_{D^d}(D\phi_i^{-1})^*\omega$
, and I kinda wanna use the fact that $\omega$ is representing the Thom class here so that this last integral is $Ind_{p_i}X$ but I am not sure that will be the case I actually think we get the area of the disk.
Any help is appreciated. Thanks in advance.