The following is theorem 5.10 of Forster's Lectures on Riemann surfaces.
Suppose $X$ is a Riemann surface and $f:X\to D^*$ is an unbranched holomorphic covering map. Then one of the following holds: (i) Skip (ii) If the covering is $k$-sheeted ($k<\infty$), then there exists a biholomorphic mapping $\varphi:X\to D^*$ such that $p_k\circ\varphi = f$, where $p_k:D^*\to D^*$ is the mapping $z\mapsto z^k$.
Here, $D^*$ is the open unit disk with the origin removed.
Proof. (ii) Since $\exp:H\to D^*$, where $H$ is the right half plane, is the universal covering, there exists a holomoprhic mapping $\psi:H\to X$ such that $\exp = f\circ\psi$. Let $G\subset Deck(H/D^*)$ be the corresponding subgroup. Then, $$Deck(H/D^*)= \{\tau_n:n\in\Bbb Z\},$$ where $\tau_n:H\to H$ denotes the translation $z\mapsto z+2\pi in$. Thus for every subgroup $G\subset Deck(H/D^*)$ which is not the identity, there exists a natural number $k\geq 1$ so that $$G = \{\tau_{nk}:n\in\Bbb Z\}.$$ Let $g:H\to D^*$ be the covering map defined by $g(z) = \exp(z/k)$. Then $g(z) = g(z')$ precisely if $z$ and $z'$ are equivalent modulo $G$. $\color\red{\text{Hence there exists a bijective mapping}\ \varphi:X\to D^*\ \text{such that}\ \varphi\circ\psi = g\ \text{holds}}$. Since $\psi$ and $g$ are locally biholomorphic, $\varphi$ is biholomoprhic. It is now easy to check that $p_k\circ\varphi =f$ and the theorem is proved.
I can't understand the highlighted red part. How can I ensure the existence of bijection $\varphi$? Also, $p_k\circ\varphi = f$ is not clear to me. I can't see why $D^*\xleftarrow{g} H\xrightarrow{\exp}D^*$ is $k$th power map. Please help.
This follows from Proposition 5.9. Both covering maps $\psi: H \to X$ and $g: H \to D^*, z \mapsto \exp(z / k)$ identify two points $x, x' \in H$ if and only if there is a deck transformation $\sigma \in G$ with $\sigma(x) = x'$. Hence both maps are canonically bijective to the quotient $$H \to H / G.$$ Hence there is a bijection $X \to D^*$, compatible with the covering maps coming from $H$.