Let $Y$ be any scheme, $S$ a set and $\underline{S}_Y$ the constant sheaf on some reasonble site over $X$, say to have something concrete in mind the small etale site $X_{\text{et}}$. Let $f: X \to Y$ any morphism of schemes. It is well known that the pullback sheaf $f^*\underline{S}_Y$ is a constant sheaf too isomorphic to $\underline{S}_X$.
There are three arguments known to me to show it:
(1) $\underline{S}_Y$ can be realized as pullback of the constant sheaf $\underline{S}_{\text{pt}}$ with respect to a map $p_Y: Y \to \text{pt}$. Then observe that $p_X^*\mathcal{F} \cong f^*p_Y^*\mathcal{F}$ for any sheaf
(2) $\underline{S}_Y$ is represenable by etale projection $Y \times S \to Y$ and the pullback is obviously representable by etale $X \times_Y (Y \times S)= X \times S$
(3) Locally constant sheaves correspond under $\mathcal{F} \mapsto \mathcal{F}_u$ for arbitrary geometric point $y: y \to Y$ to sets with $\pi_1(Y,y)$-action by algebr fundamental group. Constant sheaves correspond to those with trivial $\pi_1(Y,y)$-action. Under this equivalence the $\pi_1(X,x)$ corresponding to the pullback $f^*\underline{S}_Y$ is given by $S$ with inducted action by composition $\pi_1(X,x)\to \pi_1(Y,y)$,
But I have to admit that I still not completely not happy with these arguments, as they appear a bit too tricky or say requires relatively too much effort for this - say nearly trivial fact. Note that no one uses explicitly the standard definition of pullback sheaf as sheafification of certain presheaf; see below.
And I'm wondering if it is not possible to show this more "directly, or plainly without a roundabout way" using just directly this definition of pullback sheaf that $f^*\underline{S}_Y$ is a constant sheaf instead more or less using "tricks" or refering to deeper results.
Concern/Problem: In other words, is there a "canonical way" to prove it working only with definition of pullback sheaf obtained as sheafification of the presheaf described below? In short terms, I'm looking for a " very plain" proof of this obvious plain fact.
Recall, for any sheaf $\mathcal{F}$ on $Y$ the pullback presheaf $\text{pre} f^*\mathcal{F}$ is defined for any etale $V \to X$ by
$$ \text{pre} f^*\mathcal{F}(V)= \varinjlim_U \mathcal{F}(U) $$
where the direct limit runs over etale $U \to Y$ through which the composition $V \to X \to Y$ factors.
Now does there exist a say "straight fortward strategy" to establish that $f^*\underline{S}_Y$ is isomorphic to $\underline{S}_X$ directly using only this?
On level on stalks it's clear that $(f^*\underline{S}_Y)_{f(x)}=(f^*\underline{S}_Y)_x=S =(\underline{S}_X)_x$ and using universal property of sheafification it suffice to construct explicitly compatible maps $ \text{pre} f^*\mathcal{F}(V) \to \underline{S}_X(V) $ on level of local sections wrt etale maps $V \to X$ which gives an iso on level of stalks.
My "naive guess" is the following: Let $U \to Y$ any etale morphism in direct system defining $ \text{pre} f^*\mathcal{F}(V) $, ie $V \to X\to Y$ factors through $U$, especially $f(V) \subset U $ on level of sets/ topolog.
$V \to X$ is etale and has $n=\pi_0(V)$ components $V_1,..., V_n$ and let $U_1,..., U_m$ be the components of $U$. Then, every $f(V_i)$ by continuity is contained in a unique $U_{f{V,U}(i)}$. So we obtain a well defined function $f_{U,V}:\{1,..., n\} \to \{1,..., m\}$.
If $f_{U,V}$ is surjective, then we could constuct a map $m_U(V):\underline{S}_Y(U) \to \underline{S}_X(V)$ as follows ( and then passing to the limit over all $U$):
Let $s:=(s_1,..., s_j, ...,s_m) \in S^{\pi_0(U)}=\underline{S}_Y(U)$ with $s_j \in S $ corresponding to factor associated with $U_j$. We map this vector to the vector $t:=(t_1,..., t_k, ...,t_n) \in S^{\pi_0(V)}=\underline{S}_X(V)$ such that $t_i:= s_j$ if $j =f_{U,V}(i)$.
This should be compatible with limits so far as long as we dealing with $U \to Y$ such that $f_{U,V}$ surjective, ie every component of $U$ contains at least one component of $ f(V)$. How to extend it if that's not the case?
Modulo this problem, does this approach work? And if yes, is this the say most "canonical" approach to prove that $f^*\underline{S}_Y$ is constant using definition of pullback solely, or can be done in more sophisticated way?