By simplicial approximation, a continuous map $f:S^n \to S^n$ is homotopy equivalant to a simplicial map. Such a map must send vertices to vertices, from which the full map is determined by linearly extending it across simplices (spanned by the vertices.)
One can show that, for an interior point $q$ of a subdivison of $S^n$, that its preimage $f^{-1}(q)$ is finite, and $f$ acts as an invertible linear map in a neighborhood of these preimage points.
This is a sketch of (a) for exercise 4.1.15 of Hatcher:
I am having trouble making thorough my answer for (b).
Let $p_i \in f^{-1}(q)$, and denote $L_i'$ the neighborhood of $p_i$ on which $f$ acts invertibly and linearly. Now write $U = \cup_i f(L_i')$, and finally $L = f^{-1}(U)$.
The connected components of $L$ are just neighborhoods of the $p_i$ on which $f$ acts invertibly and linearly. Denote by $L_i$ the connected component of $L$ containing $p_i$. Define $f_i:S^n \to S^n$ as the map which acts as $f$ on $L_i$ and sends $S^n \setminus L_i$ to the basepoint.
I think $[f] = \Sigma [f_i]$. Visually, I think this is clear since we can identify $L_i$ with $S^n$, and think of quotienting out $S^n \setminus ( \cup L_i)$ as 'pinching together' portions of $f$ away from points of $f^{-1}(q)$.
But I did not really do what Hatcher said, at least, I do not see how. If I did, how does $gf$ relate? (Also, can my 'visual argument' be made clearer?)