Consider the map $f: S \to S$, such that $f: \phi \mapsto \omega \phi$. Let $\omega \in (-1,1)$. This map is not surjective, so it has $\text{deg}(f) =0$. However, if we use the degree formula
$$ \text{deg}(f) = \frac{\int_S df(\Omega)}{\int_S \Omega}$$
we get $\text{deg}(f) = \omega$. Why is that?
What about for $\omega \in \mathbb{R}$ ?
As student_du_05 pointed out, this map is not well defined unless $\omega = 0$ (then the degree formula gives $\mathop{\mathrm{deg}}(f)=0$).
The circle $S$ is defined as $\mathbb{R}$ with points $a$, $b$ identified if $a=b+2\pi n$ for some integer $n$. In order for a map $S\to S$ to be well defined, equivalence classes of points in $\mathbb{R}$ must get sent to other equivalence classes.
If the map $f$ is well defined, assuming $a=b+2\pi n$, the image of $a$ is $f(a)=\omega a= \omega b + 2\pi \omega n$. This is in the same equivalence class as $f(b)=\omega b$ if and only if $2\pi \omega n$ is an integer multiple of $2\pi$ for all $n$. This is only the case if $\omega\in\mathbb{Z}$. In such a case the degree formula will give the correct answer.