Let $F \in C^{\infty}\left(\mathbb{R}^{n}, \mathbb{R}^{n}\right)$ and $G \in C^{\infty}\left(\mathbb{R}^{m}, \mathbb{R}^{m}\right)$ be vector fields with precisely one singularity at the origin. Let $i(F,0)$ be the index of $F$ at $0$ (the index of $F$ at $0$ is the degree of $\phi$, $\phi: S^{n-1}\to S^{n-1}$, where $\phi(x)=F(x)/||F(x)||$). Similar for $i(G,0)$ and $i(F\times G,0)$. I want to prove
$$i(F\times G,0)=i(F,0)i(G,0)$$
I want to prove it by definition, but have no idea how to do it.
Step $1$: We want to find a vector field $\widetilde{F}$ on $\mathbb{R}^{n}$ with only non-degenerate singularities such that $\mathrm{Ind}(F)=\mathrm{Ind}(\widetilde{F}).$ (Here $\mathrm{Ind}(F)$ is just $i(F,0)$). For this we just need to construct a vector field which agree with $F$ outside a compact set, since the index of $F$ and $\widetilde{F}$ only depends on the restriction of them to $\partial B_{R}(0)$ for $R$ sufficiently large. Let $\beta\in C^{\infty}(\mathbb{R}^{n})$ with $\beta(x)\equiv1$ for $||x||\le1$ and $\beta(x)\equiv0$ for $||x||\ge2$. Let $$c=\mathrm{inf}_{1\le||x||\le2} ||F(x)||$$which is positive since $F$ is onle zero at $x=0$. Let $w\in\mathbb{R}^{n}$ with $||w||<c$ be a regular value of $F: \mathbb{R}^{n} \to \mathbb{R}^{n}$ (by Sard's theorem, regular values are dense in $\mathbb{R}^{n}$, so we could find this $w$). Define $$\widetilde{F}(x):=F(x)-\beta(x)w$$ Then $\widetilde{F}(x)=F(x)$, if $ ||x||\ge 2. $ $\widetilde{F}(x)\ge c-||w||>0$, if $1\le||x||\le 2$. $\widetilde{F}(x)=F(x)-w$, if $||x||\le 1$. So all zeros of $\widetilde{F}$ are in $\mathrm{int}(D^{n})$ and $\widetilde{F}^{-1}(0)=\mathrm{int}(D^{n})\cap F^{-1}(w)$. Since $D\widetilde{F}(p)=DF(p)$ and $w$ is a regular value of $F$, $D\widetilde{F}(p)$ is invertible whenever $p\in\mathrm{int}(D^{n})\cap F^{-1}(w).$ So then zeros of $\widetilde{F}$ are all non-degenerated and isolated, in particular, they are finite.
Step 2: We construct $\widetilde{G} $ similarly. Then $(p,q)$ is a singularity of $\widetilde{F}\times\widetilde{G}$ if and only if $(\widetilde{F}(p),\widetilde{G}(q))=0$ if and only if $p$ is a singularity of $\widetilde{F}$ and $q$ is a singularity of $\widetilde{G}$. Since $D(\widetilde{F}\times\widetilde{G})(p,q)=\left(\begin{array}{ll}{D\widetilde{F}(p)} & {0} \\ {0} & {D\widetilde{G}(q)}\end{array}\right)$, $\mathrm{det}D(\widetilde{F}\times\widetilde{G})(p,q)=\mathrm{det}D\widetilde{F}(p)\mathrm{det}D\widetilde{G}(q)$. Hence (p,q) is a non-degenerate singularity of $\widetilde{F}\times\widetilde{G}$ if and only if $p$ is a non-degenerate singularity of $\widetilde{F}$, $q$ is a non-degenerate singularity of $\widetilde{G}$. In this case, $i(\widetilde{F}\times\widetilde{G})(p,q)$=$\mathrm{sign}(\mathrm{det}D(\widetilde{F}\times\widetilde{G})(p,q))=\mathrm{sign}(\mathrm{det}D\widetilde{F}(p))\mathrm{sign}(\mathrm{det}D\widetilde{G}(q))=i(\widetilde{F},p)i(\widetilde{G},q)$.
Step 3: It is clear that $\mathrm{Ind}(\widetilde{F}\times\widetilde{G})=i(F\times G,0)$. So by step 1 and 2, we have $i(F,0)i(G,0)=\mathrm{Ind}(\widetilde{F})\mathrm{Ind}(\widetilde{G})=(\sum_{p}i(\widetilde{F},p))(\sum_{q}i(\widetilde{G},q))=\sum_{p,q}i(\widetilde{F},p))(\sum_{q}i(\widetilde{G},q))=\sum_{p,q}i(\widetilde{F}\times\widetilde{G},(p,q))=\mathrm{Ind}(\widetilde{F}\times\widetilde{G})=i(F\times G,0)$