Let $f:S^1\to S^1$ be a continuous map and there is $\delta>0$ such that for $x,y\in S^1$, if $\operatorname{diam} \operatorname{arc}(x,y)<\delta$, then $f$ is increasing (or decreasing) on $\operatorname{arc}(x, y)$.
What can say about $f$? Is it a local homeomorphism?
This is the "converse" of your question local homeomorphism of the circle .
As in my answer to this question we use the covering map $e : \mathbb{R} \to S^1, e(x) = e^{ix}$. Take again a lift $\tilde{f} : \mathbb{R} \to \mathbb{R}$ of $f \circ e$. W.l.o.g. let $f$ be increasing on some arc contained in $S^1$ which means that $\tilde{f}$ is increasing on some interval $[a_0 ,b_0]$. As we have shown, this implies that $\tilde{f}$ is increasing on all of $\mathbb{R}$.
Now it depends on what you understand by "increasing".
If you simply mean $\tilde{f}(x) \le \tilde{f}(y)$ for $x \le y$, then $\tilde{f}$ may be constant on some intervals (that is, $f$ may be constant on some arcs contained in $S^1$; it may even constant on all of $S^1$). In this case $f$ is not a local homeomorphism.
If you mean strictly increasing ($\tilde{f}(x) < \tilde{f}(y)$ for $x < y$), then $\tilde{f}$ must be a homeomorphism which implies that $f$ is a local homeomorphism because $e$ is one. To see this, first observe that $\tilde{f}$ is injective and an open map because it maps open intervals to open intervals. It therefore remains to show that it is surjective. This is equivalent to $\inf \tilde{f}(\mathbb{R}) = -\infty$ and $\sup \tilde{f}(\mathbb{R}) = \infty$. Consider the map $g(x) = \tilde{f}(x +2\pi) - \tilde{f}(x)$ which has the property $g(x) > 0$ for all $x$. We have $$e(g(x)) = e^{ig(x)} = e^{i\tilde{f}(x +2\pi)}/e^{i\tilde{f}(x)} = e(\tilde{f}(x +2\pi))/e(\tilde{f}(x)) = f(e(x+2\pi))/f(e(x)) = $$ $$f(e(x))/f(e(x)) = 1 .$$ This implies that $g$ takes values in $e^{-1}(1) = \{2k\pi \mid k \in \mathbb{Z} \}$ and must therefore be constant by continuity, that is $g(x) \equiv 2k\pi$ for some $k > 0$. Hence $\tilde{f}(x + 2\pi) = \tilde{f}(x) + 2k\pi$ and $\tilde{f}(x - 2\pi) = \tilde{f}(x) - 2k\pi$. By induction we get $\tilde{f}(\pm2\pi n) = \tilde{f}(0) \pm 2nk\pi$ which completes the proof.
Summarizing we can state that the following are equivalent:
(1) $f$ is a local homeomorphism
(2) $\tilde{f}$ is a local homeomorphism
(3) $\tilde{f}$ is a homeomorphism
(4) $\tilde{f}$ is strictly monotonic
(5) $\tilde{f}$ is locally strictly monotonic
(6) $\tilde{f}$ is strictly monotonic on each interval $[a,b]$ of length $b - a < 2\pi$ (this is the "correct" interpretation of $f$ being strictly monotonic on each arc in $S^1$)
(7) There exists $\tilde{\delta} > 0$ such that $\tilde{f}$ is strictly monotonic on each interval $[a,b]$ of length $b - a < \tilde{\delta}$
