I'm reading a paper about finite-order homeomorphisms of a closed oriented surface, say $f: M \rightarrow M$, $f^n = id$. I know that from such an $f$ we get a covering $P: M \rightarrow M_f$, where $M_f$ is the orbit space of $M$. (For simplicity, I am just assuming that this has no branch points.)
But then the author continually refers to "the representation of $P$", supposedly a map $\rho: \pi_1(M_f) \rightarrow \mathbb Z/n\mathbb Z$. I just do not see what this map is supposed to be, or how it is induced by $f$.
Here is an example I am trying to explicitly work through. Let $M$ be the genus-$4$ surface in the shape of a triangle, and let $f$ is a rotation of $2\pi/3$ about the center (imagine a fidget spinner..). Then $M_f$ is the genus-$2$ surface, with fundamental group $$\pi_1(M_f) = \langle a_1,b_1,a_2,b_2 \mid [a_1,b_1][a_2,b_2] = 1\rangle.$$ What is $\rho$, a map from this into $\mathbb Z/3\mathbb Z$, supposed to be?
A search through Google and Hatcher (my reference text) did not yield any definition. Please let me know if any more details would be helpful.
The key words you need are "deck transformation group" or "covering transformation group", and "regular covering space" or "normal covering space"; there's other variations on this terminology as well. If you look for the discussion on these topics in Hatcher, then you will find what you need.
Here's an outline.
Let $n$ be the least positive integer such that $f^n = \text{Id}$. Let $\langle f \rangle = \{f^i \mid i \in \mathbb Z\}$. This is a group under composition, and in fact it is a finite cyclic group generated by $f$ and isomorphic to $\mathbb Z / n \mathbb Z$, the isomorphism given by $f^i \mapsto i / n\mathbb Z \in \mathbb Z / n \mathbb Z$.
Now, you assumed that $f$ has no branch points, but let me make the required assumption a bit more explicit: What we must assume is that $\langle f \rangle$ acts freely, which means that if $i \in \{0,...,n-1\}$ and if $f^i$ has a fixed point then $i=0$. Without this assumption, the representation $\rho$ you want need not exist.
It follows from the assumption of freeness that the quotient map $P : M \to M_f$ is a regular covering map, that the deck transformation group (or covering transformation group) is the finite cyclic group $\langle f \rangle$, that the induced homomorphism $P_* : \pi_1(M) \to \pi_1(M_f)$ is injective, that its image $\text{image}(P_*) < \pi_1(M_f)$ is normal, and that the quotient group $\pi_1(M_f) \, / \, \text{image}(P_*)$ is isomorphic to the deck transformation group $\langle f \rangle$ which, as said, is isomorphic to $\mathbb Z / n \mathbb Z$. Putting this altogether, we get the desired representation $$\pi_1(M_f) \mapsto \pi_1(M_f) \, / \, \text{image}(P_*) \approx \langle f \rangle \approx \mathbb Z / n \mathbb Z $$
I could add to this a description of the example you requested, if you like, but for now perhaps you might like to work it out, knowing about the general approach.