I am reading Geordie Williamson's guide to perverse sheaves and stuck on Example 5.11.
Consider the map $f:\mathbb{C}^* \to \mathbb{C}^*: z \mapsto z^m$. Let $\underline{k}$ be the constant sheaf on $\mathbb{C}^*$ with values in $k:=\mathbb{C}$ (regarded as 1D $\mathbb{C}$-space) and we want to understand the pushforward sheaf $f_*\underline{k}$.
Obviously the stalks of $f_*\underline{k}$ are $\mathbb{C}^m$ because $f$ is a covering. In detail, let $U \subset \mathbb{C}^*$ contractible and small enough such that $f$ trivializes over $U$ to disjoint union $f^{-1}(U)= \coprod_{i=1}^m V_i$, then since $\underline{k}(U)=k$ can be identified with continuous functions $U \to k=\mathbb{C}$ where $k$ is endowed with discrete topology, then $f_*\underline{k}(U)$ is naturally identifid with set of $k$-valued continuous functions $V_1 \coprod ... \coprod V_m \to k$, which can be identified with $k^m$.
Then the part I don't understand: It is claimed that $f_*\underline{k}$ is a local system determined by the action of the monodromy on the $m$-th roots of $1$ (an $m$-cycle).
Why is that the case and how can it be verified? Secondly, the converse question is how can I construct from the local system $f_*\underline{k}$ the monodromy representation $\pi_1(\mathbb{C}^*,1) \to GL_m(k)$ in explicit terms?
I know how to construct explicitly from a given monodromy representation $\pi_1(\mathbb{C}^*,1) \to GL_n(k)$ a local system (the construction is presented e.g. here ). But if not provides an explicit construction in another direction.
Given a local system $\mathcal{L}$ on a path connected space $X$ and a point $x$, you can define the corresponding monodromy on $\mathcal{L}_x$ as follows : take a path $\gamma : [0,1]\to X$ such that $\gamma(0)=\gamma(1)=x$. Then take the pullback $\gamma^{-1}\mathcal{L}$. Since $[0,1]$ is a contractible space, $\gamma^{-1}\mathcal{L}$ is trivial and thus it induces for any $t \in [0,1]$ the map $\Gamma([0,1],\gamma^{-1}\mathcal{L})\to (\gamma^{-1}\mathcal{L})_t$ is an isomorphism.
The monodromy is the map $$\mathcal{L}_x = (\gamma^{-1}\mathcal{L})_0 \overset{\sim}\longleftarrow \Gamma([0,1],\gamma^{-1}\mathcal{L}) \overset{\sim}\longrightarrow (\gamma^{-1}\mathcal{L})_1 = \mathcal{L}_x$$ You can show that it only depends on the homotopy class of $\gamma$ in $\pi_1(X,x)$ and that it is indeed a well defined action of $\pi_1(X,x)$.
Concretely, let's have a look at the local system $f_*\underline{k}$ where $f:\mathbb{C}^*\to\mathbb{C}^*$ is the $m$-fold covering $z\mapsto z^m$. Let $\gamma:[0,1]\to \mathbb{C}^*$ be the map $t\mapsto e^{2i\pi t}$. The path $\gamma$ has $m$ liftings $\gamma_0,\gamma_1,...,\gamma_{m-1}$ where $\gamma_n:[0,1]\to\mathbb{C}^*$ is the map $t\mapsto e^{2i\pi(t+n)/m}$. The main ingredient is the fact that $\gamma_n(0)=\gamma_{n-1}(1)$ for all $n\in\mathbb{Z}/m\mathbb{Z}$.
Before a more precise look, let's introduce some notations. We will write $\eta = e^{2i\pi/m}$. So we have $\eta^{n}=\gamma_n(0)=\gamma_{n-1}(1)$. And to avoid confusion between the point $1\in[0,1]$ and $1\in\mathbb{C}^*$, I will now on write the latter $x=\gamma(0)=\gamma(1)$.
By definition, $f_*\underline{k}$ is the sheaf such that section on $U$ are locally constant function on $f^{-1}(U)$ with value in your coefficient ring $k=\mathbb{C}$. As you said, the stalk at $x$ is $(f_*\underline{k})_x = k ^m$ where on the $n$-th coordinate (starting at 0) is the value of the locally constant function at $\eta^n=e^{2i\pi n/m}$. So maybe, it is best to write $$(f_*\underline{k})_x = \operatorname{Maps}(f^{-1}(x),k) = \operatorname{Maps}(\{\eta^0, \eta^1,...,\eta^{m-1}\}, k)\simeq k^m$$
Now, the pullback $\gamma^{-1}f_*\underline{k}$ is the sheaf of tuples of $m$ functions $(y_0, y_1, ..., y_{m-1})$ where each $y_n$ is defined on $[0,1]$ and takes a locally constant value in $k$. But we need to think of each $y_n$ as a locally constant function defined on the image of the path $\gamma_n$.
The stalk at $t\in[0,1]$ is their values, namely $(y_0, y_1, ..., y_{m-1})_t = (y_0(t),...,y_{m-1}(t))$. But now, to make the identification $$(\gamma^{-1}f_*\underline{k})_t = (f_*\underline{k})_{\gamma(t)}$$ we need to remember that each $y_n$ is correspond to a locally constant function on the image of $\gamma_n$. Thus, given a $(y_0,...,y_{m-1})_t \in (\gamma^{-1}f_*\underline{k})_t$, its image in $(f_*\underline{k})_{\gamma(t)}$ is the map $f^{-1}(\gamma(t))\to k$ such that $\gamma_n(t)\mapsto y_n(t)$.
It follows that the two maps of interest are : $$\begin{array}{rcccl} \Gamma([0,1],\gamma^{-1}f_*\underline{k})&\longrightarrow&(\gamma^{-1}f_*\underline{k})_0&\longrightarrow&(f_*\underline{k})_x\\ (y_0,...,y_n)&\longmapsto&(y_0(0),...,y_{m-1}(0))&\longmapsto&(\eta^n=\gamma_n(0)\mapsto y_n(0))\end{array} $$ and $$\begin{array}{rcccl} \Gamma([0,1],\gamma^{-1}f_*\underline{k})&\longrightarrow&(\gamma^{-1}f_*\underline{k})_1&\longrightarrow&(f_*\underline{k})_x\\ (y_0,...,y_n)&\longmapsto&(y_0(1),...,y_{m-1}(1))&\longmapsto&(\eta^n=\gamma_{n-1}(1)\mapsto y_{n-1}(1))\end{array} $$ Of course, each function $y_n$ is actually constant so $y_n(0)=y_n(1)$.
Hence, the monodromy $$(f_x\underline{k})_x = (\gamma^{-1}f_x\underline{k})_0 \overset{\sim}\longleftarrow \Gamma([0,1],\gamma^{-1}f_x\underline{k}) \overset{\sim}\longrightarrow (\gamma^{-1}f_x\underline{k})_1 = (f_x\underline{k})_x$$ is a cycle. If we write $(f_*\underline{k})_x\simeq k^m$ as a set of tuples where on the $n$-th coordinate (starting at 0) is the image of $\eta^n$, then the monodromy is $(y_0,y_1,...,y_{m-1})\mapsto (y_{m-1},y_0,y_1,...y_{m-2})$.