Consider the double cover $\pi:S^1 \rightarrow S^1, z \mapsto z^2$ and the pushforward of the constant sheaf $\pi_{*}\mathbb{Z}$. This is a locally constant sheaf of rank 2, but not constant (since the space of global sections is rank 1).
Question: if I choose a basis $u,v$ for the stalk at some point $p$, how to compute the monodromy matrix with respect to this matrix?
It looks like this was asked and answered a while ago. But since it floated to the top again, let me give another answer.
First in classical language, the monodromy exchanges sheets $ \sqrt{z}\leftrightarrow -\sqrt{z}$. If you imagine taking formal linear combinations of these, then you should be able to recover the monodromy matrix given in Francesco's answer.
Here is a more abstract general point of view. If $Y$ is a connected (sufficiently nice) space, the category of locally constant sheaves is equivalent to representations of $\pi_1(Y)$ via monodromy. The pullback of a locally constant sheaf along a covering space $\pi:X\to Y$ corresponds to restriction from $\pi_1(Y)$ to $\pi_1(X)$. Pushforward would be the right adjoint which corresponds to induction in the opposite direction. So in particular $\pi_*\mathbb{Z}$ is the regular representation of $\pi_1(Y)\to Aut(\mathbb{Z}[G])$ when $\pi$ is Galois with group $G$.
Let $w \in S^1$, and let $\pi^{-1}(w)=\{z_1, z_2 \}$.
Let $U$ be a sufficiently small open set containing $w$, so that $\pi^{-1}(U)$ is the disjoint union of two open sets $V_1$ and $V_2$, with $V_i$ containing $Z_i$.
Then, by definition, the space of sections of $\pi_* \mathbb{Z}$ over $U$ is the direct sum of the spaces of sections of $\mathbb{Z}$ over $V_1$ and $V_2$. So we are reduced to compute the monodromy action of the fundamental group of $S^1$ on the fibre $\pi^{-1}(w)$.
Let us fix the point $1$ as a base point for $\pi_1(S^1) \cong \mathbb{Z}$. When we consider the action of the generator $1 \in \pi_1(S^1)$ on $\pi^{-1}(w)$, one easily checks that $z_1 \to z_2$ and $z_2 \to z_1$.
Hence the monodromy matrix is $$\left(\begin{matrix}0 & 1 \cr 1 & 0 \end{matrix}\right).$$
