Degree of a map $f: S^1 \to S^1$. I’m reading the following definition.

Definition B11.1. Given a map $f: S^1 \rightarrow S^1$, we define its degree $\operatorname{deg} f \in \mathbb{Z}$ as follows. Choosing a basepoint $x_{0} \in \mathbb{S}^{1},$ we get
$$
f_{*}: \mathbb{Z} \cong \pi_{1}(\mathbb{S}^{1}, x_{0}) \mapsto \pi_{1}(\mathbb{S}^{1}, f(x_{0})) \cong \mathbb{Z}
$$
Any such homomorphism $\mathbb{Z} \rightarrow \mathbb{Z}$ is given as multiplication by some $n \in \mathbb{Z}$, meaning the map $k \mapsto n k$. We then set $\operatorname{deg} f:=n$

While in our class we’ve covered why $\mathbb{Z} \cong \pi_{1}(\mathbb{S}^{1}, x_{0})$, I’m not sure I get how that translates into equivalency between the maps $\pi_{1}(\mathbb{S}^{1}, x_{0}) \to \pi_{1}(\mathbb{S}^{1}, f(x_{0}))$ and the maps $S^1 \to S^1$. Say we embed $S^1$ into $\mathbb{R}^2$, and $f$ rotates $x_0 = (1,0)$ by an angle of $\pi/3$ around the origin. What would $n$ be in this case? And given $k \mapsto nk$, what does it mean in terms of the homotopy classes of loops in $S^1$? If it helps please use concrete numbers for $n$ and $k$.
Edit: It’s been pointed out to me that orientation matters. Indeed this comes right after the definition in my materials. For now, I’m just assuming we map the loop going counter-clockwise around $S^1$ once to $1$.
 A: It seems as though the degree is well-defined only up to a sign. The reason I say this is because while it is true that $\pi_1(\Bbb S^1, f(x_0)) \cong \Bbb Z$, there are two isomorphisms, depending on which generator you pick.
In your example, you technically haven't given a map $f: \Bbb S^1 \to \Bbb S^1$, you've only told what $f$ does to $(1, 0)$. I assume that you mean that $f$ rotates $\Bbb S^1$ by $\pi/3$. In this case, the degree could either be $+1$ or $-1$.
To get a geometric interpretation: Recall what the isomorphism $G = \pi_1(\Bbb S^1, x_0) \cong \Bbb Z$ explicitly is. There are exactly two generators of $G$. One is (the equivalence class of) the loop that goes around exactly once and the other is the reverse of that. We get the isomorphism by mapping one of them to $1$. This means that the degree of the map is simply measuring how many times the generating loop gets twisted, under $f$. The only problem is that you have a choice in fixing the isomorphism, which means that the orientation matters.
