Let $\psi$ be a homeomorphism and $\gamma :[0,1] \rightarrow \mathbb{R}^2$ a path. Now assumme additionally that $(\psi \circ \gamma)(t) \neq \gamma(t)$ everywhere. Then we can look at the map $\rho_{\gamma}(t) = \frac{(\psi \circ \gamma)(t) - \gamma(t)}{|(\psi \circ \gamma)(t) - \gamma(t)|} \in S^1.$
Now, such a map can also be parametrized as $\rho_{\gamma}(t) = e^{i \alpha_{\gamma}(t)},$ where $\alpha_{\gamma}(t) \in [0,2\pi).$
My question is now: Imagine that $\gamma_{0}$ and $\gamma_{1}$ are homotopic paths, how can I understand that $\alpha_{\gamma_{0}}$ and $\alpha_{\gamma_{1}}$ must be homotopic, too?
I guess it has something to do with liftings, but I don't see quite through the details.
$\alpha_\gamma(t)$ just represents the angle made by the vector drawn from $\gamma(t)$ to $\psi(\gamma(t))$, which you could write down an explicit formula for if you wanted to.
Although it is true that $\alpha_\gamma(t)\in[0,2\pi)$, if you define it this way, it won't be continuous since it will in general jump from $2\pi$ back to $0$. You could let $\alpha_\gamma(t)$ take values in $\mathbb R$ to get around this, although all paths are homotopic in $\mathbb R$ since it is contractible. (If you want to also fix the endpoints, then you need to assume that $\alpha_{\gamma_0}$ has the same endpoints as $\alpha_{\gamma_1}$ which it won't in general.
I think the most natural thing is simply to think of $\rho_{\gamma}(t)$ as an element of the circle. Then it makes sense to ask whether $\rho_{\gamma_0}$ is homotopic to $\rho_{\gamma_1}$ for any pair of initial arcs.