It is well known that over a topological space $X$ (and choosing an open cover $\mathfrak{U}$) every locally constant Cech cocycle $g$ on $\mathfrak{U}$ with coefficients in a group $G$ yields a $G$-covering space $X_g \rightarrow X$. As such, the monodromy action of this covering space gives a homomorphism from the fundamental group $\pi_1(X,x)$ on a point $x\in X$ to $G$.

I am trying to write this homomorphism explicitly in terms of the cocycle $g$. In his Bachelor Thesis, Lemma 5.5, M.P. Noordman claims that this can be done in the following way. You consider a loop $\sigma:[0,1]\rightarrow X$ and apply the Lebesgue number lemma to get a finite subcover $\{U_1,...,U_n\}$ of $\mathfrak{U}$ and a partition $t_0<t_1<...<t_n$ of the interval $[0,1]$ in such a way that $\sigma([t_{i-1},t_i])\subset U_i$. Now, you can define the homomorphism $f:\pi_1(X,x)\rightarrow G$ as $$ f([\sigma])=g_{12} g_{23} g_{34} \cdots g_{(n-1) n}. $$

However, it is not clear to me why this does not depend on the choice of the "Lebesgue subcover" $\{U_1,...,U_n\}$ or on the choice of the representative of the class $[\sigma]$.

For example, consider the case where $\mathfrak{U}=\{U,V,W\}$ consists on three open sets with $U\cap V \neq \varnothing$ and $U,V \subset W$. If we choose a path contained in $U\cup V$, we could choose the Lebesgue covering to be $\{U,V\}$, which would yield $f([\sigma])=g_{UV}$ or we could choose the covering to be simply $\{W\}$, which would yield $f([\sigma])=1$, and I do not see why these should coincide.