Let $f:S^n\to S^n$ be a continuous map. Let $x\in S^n$ and $y=f(x)$. Assume that there is a neighborhood $U$ of $x$ in $S^n$ such that no element of $U-\{x\}$ maps to $y$.
Question. Can the interior of $f(U)$ be empty?
Motivation. (I made an error in defining the map $\bar f$ (see below), so you may skip this part. Thanks to @EricWofsey for pointing out the mistake).
If the answer to the above question is no, then we can attempt to define the local degree of $f$ in a geometric way.
For then we can do the following: Assume that $U$ is "nice". That is, $U$ is the intersection of $S^n$ with a small ball $B(x, \varepsilon)\subseteq \mathbf R^{n+1}$. Thus $S^n/(S^n-U)$ is homeomorphic to $S^n$.
Now let $V$ be a "nice" neighborhood of $y$ contained in $f(U)$. We get a map $$\bar f: S^n/(S^n-U)\to S^n/(S^n-V)$$ induced from $f$. Both the domain and target spaces of $\bar f$ are homeomorphic to $S^n$ so we can talk about the degree of $\bar f$ (provided we have a scheme to consistently choose generators of $H_n(S^n/(S^n-U))$ and $H_n(S^n/(S^n-V))$).
Yes, this can happen. For instance, consider the map $f:\mathbb{R}^n\to\mathbb{R}^n$ given by $f(x)=(\|x\|,0,\dots,0)$ (where $\|x\|$ is the Euclidean norm), which extends continuously to a map $S^n\to S^n$ between the one-point compactifications. For $x=(0,0,0,\dots,0)$, there are no preimages of $f(x)$ besides $x$, but $f(S^n)$ has empty interior if $n>1$.
(For $n=1$, it is easy to see that $f(U)$ must have nonempty interior as long as $f$ is not constant on $U$.)