The local index is additive under a transverse perturbation

Choose a smooth parametrization $b:D^n\to B$ and pull back the field as $Y(u)=(d{b}_u{)}^{-1}X(b(u))$ on $D:=D^n$: this is a smooth field on the compact manifold $D$ with boundary, with the single zero $b^{-1}(p)$ and $Y\mathrm{\ne }0$ on $\mathrm{\partial }D$; [F2] applied to $N=D$ gives ${\mathrm{ind}}_{b^{-1}(p)}Y=\mathrm{deg}(\mathrm{\partial }D\to S^{n-1},y\mapsto Y(y)\mathrm{/}\mathrm{\mid }Y(y)\mathrm{\mid })$, and this degree is the degree of $x\mapsto X(x)\mathrm{/}\mathrm{\mid }X(x)\mathrm{\mid }$ on $\mathrm{\partial }B$ in the induced trivialization; any other orientation-compatible trivialization has the same degree by the matrix-contraction argument of [F1]; hence (i).

For (ii), the same computation applies to the transported field $Y^{\mathrm{\prime }}$ on $D$: its zeros in $\mathrm{int}D$ are $b^{-1}(q_1),\dots ,b^{-1}(q_k)$ and it agrees with $Y$ on $\mathrm{\partial }D$, so [F2] gives ${\sum }_i{\mathrm{ind}}_{b^{-1}(q_i)}{Y}^{\mathrm{\prime }}=\mathrm{deg}(\mathrm{\partial }D,Y\mathrm{/}\mathrm{\mid }Y\mathrm{\mid })={\mathrm{ind}}_{p}X$ by step 1.1; the indices are transported back by [F1], and each equals $\pm 1$ by [F3].

For (iii), let $B$ be an arbitrarily small embedded closed ball around $p$ with no other zero of $X$ in its interior and let $\phi$ be a chart on a neighbourhood of $B$ with $\phi (p)=0$; choose a smooth radial bump $\lambda$ equal to $1$ on a ball $B_1\subset B$ around $p$ and supported in a slightly larger ball $B_2$ with $B_1\subseteq B_2$, $\overline{B_2}\subseteq \mathrm{int}B$, and $B_2$ containing no zero of $X$ except $p$, and by [F4] choose a regular value $y$ of $X_{\phi }$ with $\mathrm{\mid }y\mathrm{\mid }$ smaller than the (positive) minimum of $\mathrm{\mid }{X}_{\phi }\mathrm{\mid }$ on the compact collar $\mathrm{supp}\lambda \setminus B_1$. Then $X^{\mathrm{\prime }}:=X-\lambda y$ (read in the chart) equals $X$ outside $\mathrm{supp}\lambda \subseteq \mathrm{int}B$, is nowhere zero on the collar $\mathrm{supp}\lambda \setminus B_1$, and on $B_1$ has the zeros $(X_{\phi }{)}^{-1}(y)$, a finite set of nondegenerate points because $y$ is a regular value; all these zeros lie in $\mathrm{int}{B}_2\subseteq \mathrm{int}B$, so (ii) applies and gives ${\sum }_i{\mathrm{ind}}_{q_i}{X}^{\mathrm{\prime }}={\mathrm{ind}}_{p}X$, with each index $\pm 1$ by [F3].