Let $S^n$ be the sphere of dimension $n$. In order to construct a map $f:S^n\rightarrow S^n$ of degree $d\geq 2$ one has the following construction: Let $K$ be the complement of $d$ disjoint n-dimensional open discs in $S^n$. Then collapsing $K$ to a point we find that $S^n/K\simeq \bigvee _{i=1}^d S^n$. Identifying (choosing a homemomorphism) each of the sphere in the disjoint union with $S^n$ (with the approriate orientation), the composition of the two maps $$ S^n\rightarrow \bigvee _{i=1}^d S^n \rightarrow S^n, $$ gives us a map $\phi_d:S^{n}\rightarrow S^n$ of degree $d$.

In general, if $f:S^n\rightarrow S^n$ is a map of degree $d$ and $x\in S^{n}$ is such that the fiber $f^{-1}(x)$ is **finite**, then one has from excision theorem that
$$
\sum_{y\in f^{-1}(x)} deg_f(y)=d.
$$

Q1: How would you construct a (continuous) map $f:S^n\rightarrow S^n$ of degree $d$ such that on a **dense subset of points** $X\subseteq S^{n}$ one has that for every $x\in X$ that the fiber $f^{-1}(x)$ is infinite?

Q2: Having a map $f$ as in Q1 and a point $x\in X$, is it possible to take some kind **natural** average sum over the local degrees of the elements of $f^{-1}(x)$ in such a way that the sum converges to $d$ (you may assume that $S^n$ is endowed with a metric if you think it helps) ?

**added**: Note that if one constructs a map $f:S^n\rightarrow S^n$ of degree one with infinite fibers (on a dense set) then by post-composing with a map of degree $d$ we obtain a map of degree $d$ which satisfies all the conditions.

somepoint $x$ have an infinite fiber. The example you gave at the beginning of the question already does that. The set $K$ that you collapse to a point is one of the fibers of your map. – Andreas Blass Oct 29 '12 at 0:35