Proof:

Since $f({\bf x}) \neq g({\bf x})$ the reader may verify that $tAf({\bf x}) + (1-t)g({\bf x}) \neq 0$ for any $t \in [0, 1]$. Normalizing we get the desired homotopy:

$$F:(t, {\bf x})\mapsto \frac{tAf({\bf x}) + (1-t)g({\bf x})}{\Vert tAf({\bf x}) + (1-t)g({\bf x})\Vert}, \quad t \in [0, 1], \;\; {\bf x} \in S^n.$$