I'll first outline a bit of the calculation and then deal with the statement in bold.We have the universal covering map $\pi : \mathbb{R} \to S^{1} \simeq \mathbb{R}/\mathbb{Z}$. Let $\{U_{0},U_{1},U_{2} \}$ be the good cover given in the picture above.
We have that $C^{0}(\mathfrak{U},\mathfrak{H}^{0})=\mathbb{R}^{\mathbb{Z}} \oplus \mathbb{R}^{\mathbb{Z}} \oplus \mathbb{R}^{\mathbb{Z}}$ where $\phi_{0},\phi_{1},\phi_{2} : \mathbb{Z} \to \mathbb{R}$ are arbitrary components of each $\mathbb{R}^{\mathbb{Z}}$ component defined by values in each component/interval of $\pi^{-1}(U_{i})$ for each $i=0,1,2$.
Similarly, $C^{1}(\mathfrak{U},\mathfrak{H}^{0})= \mathbb{R}^{\mathbb{Z}} \oplus \mathbb{R}^{\mathbb{Z}} \oplus \mathbb{R}^{\mathbb{Z}}$ corresponding to $U_{0} \cap U_{1},U_{0} \cap U_{2},U_{1} \cap U_{2}$
The boundary operator $\delta : C^{0}(\mathfrak{U},\mathfrak{H}^{0}) \to C^{1}(\mathfrak{U},\mathfrak{H}^{0})$ is then given by $\delta(\phi_{0},\phi_{1},\phi_{1}):=(f_{0,1},f_{0,2},f_{1,2})$ where the $f_{i}$ are defined by:
$f_{0,1}(n):=\phi_{1}(n)-\phi_{0}(n),f_{0,2}(n):=\phi_{2}(n)-\phi_{0}(n-1),f_{1,2}(n):=\phi_{2}(n)-\phi_{1}(n)$
Try and see why this is the case from the definition of the boundary operator. Next, I leave it as an exercise to show the surjectivity of the $\delta$.
$Ker(\delta)$ is clearly just $\{(\phi_{0},\phi_{1},\phi_{2}) | \phi_{0}=\phi_{1}=\phi_{2}=r\}$ for some constant $r\in \mathbb{R}$. So, $Ker(\delta) \simeq \mathbb{R}$.
Now, clearly $H^{i}(\mathfrak{U},\mathfrak{H}^{0}) = \mathbb{R}$ when $i=0$ and $0$ otherwise.
Also, note that actually $\mathbb{R}^{\mathbb{Z}} \oplus \mathbb{R}^{\mathbb{Z}} \oplus \mathbb{R}^{\mathbb{Z}} \simeq \mathbb{R}^{\mathbb{Z}}$
Best Answer
One of the lemmas about fiber bundles is that the restriction of every bundle to a contractible subspace is trivial.
Let's choose four points $p,q,r,s$ equally spaced around $S^1$, and $p$ will be the base point. Since $S^1-\{q\}$ is contractible, using the trivialization over it we obtain a homeomorphism $\mu : \pi^{-1}(p) \to \pi^{-1}(r)$. And since $S^1 - \{r\}$ is contractible, using the trivialization over it we obtain a homeomorphism $\nu : \pi^{-1}(r) \to \pi^{-1}(p)$. By composition we obtain the homeomorphism $$f : F = \pi^{-1}(p) \xrightarrow{\mu} \pi^{-1}(r) \xrightarrow{\nu} \pi^{-1}(p) = F $$ This homeomorphism $f : F \to F$ is by definition, the monodromy map (there's something to prove, namely that it is well-defined up to homotopy).
And now once you have it, you can start looking at what it induces. For instance $f$ induces by the usual pullback process the isomorphism $f^* : H^k(F;\mathbb C) \to H^k(F;\mathbb C)$.
Added after some comments: Here's a more general point of view which works when the base space $S^1$ is replaced by a more arbitrary space $B$. One should assume that $B$ is path connected, and perhaps also that $B$ is a reasonable nice kind of space such as a manifold or a CW complex. As indicated in the comments, in the case $B=S^1$ this definition is equivalent to the one given above, but it is longer and more complicated; nonetheless, as long as one understands both definitions, the proof of equivalence should be easy.
Pick a base point $p \in B$. Choose $F \subset M$ to be the "base fiber" $\pi^{-1}(p)$.
One associates to the bundle a "monodromy homomorphism", which one can think of as a group homomorphism whose domain is the fundamental group $\pi_1(B,p)$. The target group of this homomorphism can be thought of as a "mapping class group" of the fiber $F$. For instance, if the only structure on $F$ is a topology then this is the group of self-homeomorphisms of $F$ modulo the normal subgroup of homomorphisms that are isotopic to the identity. (With more structure around, one can restrict to mapping class groups that respect that structure).
Here's how the monodromy homomorphism is defined. Consider a closed path $\gamma : [0,1] \to B$ based at $p$. Let $\gamma^* M$ denote the pullback bundle: it is the subset of $[0,1] \times M$ consising of all $(t,x)$ such that $\gamma(t) = \pi(x)$, and the projection $\gamma^*M \mapsto [0,1]$ defines a bundle with fiber $F$. Since the base space is contractible, this bundle is trivial. Pick a trivialization $\tau : [0,1] \times F \to \gamma^*M$ which has the property that that the map $$F \xrightarrow{x \to (0,x)} \{0\} \times F \xrightarrow{\tau} \gamma^* M \xrightarrow{\text{projection to $M$}} F $$ is the identity. The value of the monodromy map on the element of $\pi_1(B,p)$ represented by $\gamma$ is the element of the mapping class group of $F$ that corresponds to the homeomorphism $$F \xrightarrow{x \to (1,x)} \{1\} \times F \xrightarrow{\tau} \gamma^* M \xrightarrow{\text{projection to $M$}} F $$