Locally, $\mathcal A$ is a constant sheaf with stalks $\mathbb Z \oplus\mathbb Z$.
More precisely, if $U$is an open of $S^1$, $\underline {\mathbb Z}(U)$ is the product of one copy of $\mathbb Z$ for each connected component of $U$, and if $V$ is an open of $\mathbb P^1$, then $\mathcal A (V) = \underline{\mathbb Z}(f^{-1} (V))$ = the product of as many copies of $\mathbb Z$ as there are connected components of $f^{-1} (V)$, which is usually twice the number of connected components of $V$.
For example, if we take two distinct points $x_1,x_2 \in \mathbb P^1$, and take $V_i = \mathbb P^1 - \{x_i\}$, then $\mathcal A |_{V_i}$ is a constant sheaf with stalks $\mathbb Z \times\mathbb Z$. (the connected components of $f^{-1}(U)$ are always twice the connected components of $U$ when $U$ is an open subset of $V_i$)
But, as a whole, $\mathcal A$ is not the constant sheaf on $\mathbb P^1$, because $\mathcal A(\mathbb P^1) = \underline {\mathbb Z} (S^1) = \mathbb{Z}$, which is not $\mathbb Z \oplus \mathbb Z$.
We can describe it instead as a locally constant sheaf, which is constant on $V_1$ and $V_2$, and we need to describe how it glues.
Let $W = V_1 \cap V_2$.$W$ has two components, call them $W = W_a \cup W_b$. The fact that $\mathcal A$ is constant on $V_i$ means that it gives two different isomorphisms from $\mathcal A(W_a) \to \mathcal A(W_b)$, one going through $V_1$ and the other going through $V_2$.
Denote $W_a^1, W_a^2$ the connected components of $f^{-1}(W_a)$, do the same for $W_b^1, W_b^2$, and put the exponents such that $W_a^1$ and $W_b^1$ are in the same connected component of $f^{-1}(V_1)$, so that the connected components of $f^{-1}(V_1)$ correspond to $W_a^1 \cup W_b^1$ and $W_a^2 \cup W_b^2$, and those of $f^{-1}(V_2)$ correspond to $W_a^1 \cup W_b^2$ and $W_a^2 \cup W_b^1$
Then, we have restriction isomorphisms $\rho_i^a : \mathcal A(V_i) \to \mathcal A(W_a)$, which look like this : $\rho_1^a(x,y) = (x,y)$ and $\rho_2^a(x,y) = (x,y)$ ; and $\rho_i^b : \mathcal A(V_i) \to \mathcal A(W_b)$ : $\rho_1^a(x,y) = (x,y)$ and $\rho_2^a(x,y) = (y,x)$.
Then, the twisting map is the isomorphism $\tau = \rho_2^a \circ (\rho_2^b)^{-1} \circ \rho_1^b \circ (\rho_1^a)^{-1} : \mathcal A(W_a) \to \mathcal A(W_a) $. You should get that $\tau(x,y) = (y,x)$. This isomorphism describes what happens to sections on $W_a$ when you go once through $\mathbb P^1$ : the two connected components of $f^{-1}(W_a)$ get switched around, and you are basically right when you describe it in your comment.
And in fact, this map is enough to recover all the information you need to describe the sheaf $\mathcal A$ completely, so we can describe $\mathcal A$ as a locally constant sheaf whose stalks are $\mathbb Z \oplus \mathbb Z$ twisted by $\tau$ when we do one loop around $\mathbb P ^1$