Consider the smooth manifold $C\subset\mathbb{R}^3$ and the vector field $V$ given by$$C=\{(x,y,z)\in\mathbb{R}^3|x^2+y^2=1\}$$$$V_{(x,y,z)}=\begin{pmatrix}x^2-1\\xy\\xz\end{pmatrix}$$ denote by $V_C$ the restriction of $V$ to $C$.
With some trivial calculations we can verify that indeed $V_C$ is tangent to $C$. We can observe now that $V_C$ is zero only in $(\pm 1,0,0)$.
We wish now to compute the index of $V_C$ around $p=(1,0,0)$. Consider the local parametrization around $p$ given by$$g:(-1,1)\times\mathbb{R}\rightarrow C$$$$g(u,v)=(\sqrt{1-u^2},u,v)$$ Thus $g(O)=p$ and we have:$$dg=\begin{pmatrix}\frac{-u}{\sqrt{1-u^2}} &0\\1 & 0\\ 0 & 1\end{pmatrix}$$$$(V_C)_{g(u,v)}=\begin{pmatrix}-u^2\\u\sqrt{1-u^2}\\v\sqrt{1-u^2}\end{pmatrix}$$ from here follows:$$g^*V_C=\sqrt{1-u^2}\begin{pmatrix}u\\v\end{pmatrix}$$
The map $\widehat{g^*V_C}:\partial B(O,\frac{1}{2})\rightarrow S^1$ defined as $\frac{g^*V_C}{||g^*V_C||}$ is the multiplication by $2$ and is thus an orientation preserving diffeomorphism, it follows that the sum of the indeces of isolated zeroes of $g^*V_C$ inside $B(O,\frac{1}{2})$ is $1$. Since $O$ is the only zero in this ball it follows that $g^*V_C$ has index $1$ in $O$.
I'm having a hard time following the argument after the expression of $V_C$ restricted to the image of $g$ is given. How is the expression for $g^*V_C$ obtained?
Also, is there any particular reason to consider the normalized field along the border of the ball with radius $\frac{1}{2}$ other than that it contains only the isolated zero in the origin? Once we have that, it follows that this map is the "multiplication by $2$" by computing the norm of $g^*V_C$ which comes out to be $\frac{1}{2}\sqrt{1-u^2}$, thus we could have considered this same map on the border of the ball of radius $1/n$ with $n\ge 2$ obtaining as normalized field a map which would be the multiplication by $n$ and concluded with the same argument, is this correct?
Let $G:=g((-1,1)\times\mathbb{R})\subset C$, then we wish to find the map $g^{-1}:G\rightarrow (-1,1)\times\mathbb{R}$
This is an easy matter as, given $(x,y,z)^T\in G$ we have$$\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}\sqrt{1-u^2}\\u\\v\end{pmatrix}\implies y=u,\;\;z=v$$ hence we get$$g^{-1}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}y\\z\end{pmatrix}$$ And it is trivial that $(y,z)^T\in(-1,1)\times\mathbb{R}$. Therefore we obtain$$dg^{-1}=\begin{pmatrix}0&1&0\\0&0&1\end{pmatrix}$$ from which$$g^*V_C=dg^{-1}_{g(u,v)}\circ (V_C)_{g(u,v)}$$ that is$$g^*V_C=\begin{pmatrix}0&1&0\\0&0&1\end{pmatrix}\begin{pmatrix}-u^2\\u\sqrt{1-u^2}\\v\sqrt{1-u^2}\end{pmatrix}=\sqrt{1-u^2}\begin{pmatrix}u\\v\end{pmatrix}$$ which is what we wanted.
As for the second question, it is a known fact that the index of a vector field around a zero does not depend on the choice of radius $\epsilon$ as long as the zero around which we're calculating the index is the only one contained in the image of the ball $B(0,\epsilon)$.