Embedding of Picard functor into $\text{Hom}_k(-,\text{Pic}(X/k))$ $\DeclareMathOperator\Hom{Hom}\DeclareMathOperator\Pic{Pic}\DeclareMathOperator\Spec{Spec}\DeclareMathOperator\pr{pr}$
Let $X$ be an algebraic variety $X$, that is proper over $k$ (here a variety is a scheme $X/k$ such that$\overline{X}= X \times \Spec(\overline{k})$ is irreducible
and reduced).
We consider the Picard functor  given for any $k$ scheme $S$ by
$$ \mathcal{Pic}_{X/k}(S) :=    \\
 \{ \mathcal{M} \text{ invertible sheaf 
on } X \times_k S \} / \{ \text{ inv. sheaves of the form }
p^*_S(\mathcal{K}) \text{ for } \mathcal{K} \text{ invertible on }
S \}. $$
It is known that this functor is not always representable, but almost;
that means precisely there exists a $k$-scheme
$\Pic(X/k)$ representing the associated
functor $\text{Hom}( \ , \Pic(X/k))$ which contains the
Picard functor  $\mathcal{Pic}_{X/k}$
in the sense that for any $k$-scheme $S$ there is a functorial
inclusion
$$ \iota_S: \mathcal{Pic}_{X/k}(S) \hookrightarrow \Hom_k(S,\Pic(X/k)).  $$
In general that's a proper inclusion.
The equality only holds if $X \times_k S$ admits a section over $S$.
My question if there is a way to write down explicitly the map $\iota_S$, ie given a class $[\mathcal{L}]$ of an invertible sheaf $\mathcal{L}$ on $X \times S$, what is the morphism $\iota_S([\mathcal{L}]): S \to \Pic(X/k)$ on $k$-rational points of $S$?
The most natural way to associate to $[\mathcal{L}]$ a map from $S$ to the Picard group on level of $k$ rational point is perhaps via
$s \mapsto [\mathcal{L} \vert _{X \times s}]$. This is of couse welldefined and the construction is so "canonical", that it suggests that's the only possible way how $\iota_S$ could like.
But I can't find a strictly formal argument why $\iota_S$ is defined in that way? The question is closely related to this one
 A: $\DeclareMathOperator\Hom{Hom}\DeclareMathOperator\Pic{Pic}\DeclareMathOperator\Spec{Spec}\DeclareMathOperator\pr{pr}$
One easily observes that  $\forall s \in S, \mathcal{Pic}_{X/k}(s) = \Pic(X/k)(k)$, $\Hom(s, \Pic(X/k)) = \Pic(X/k)(k)$, where $Y(k)$ is the set of $k$-points of $Y$. We assume that forall $s \in S$, the morphism $i_s$ is the identity. This is a normalization condition for the homomorphims of functors $\iota$. It could be chosen otherwise, but this one seems the most natural.
Let $s \in S$ be a rational point and consider $\sigma_s : s \longrightarrow S$ the corresponding section. Since $\iota$ is a homomorphism of functors, we have a commutative diagram:
$$
\require{AMScd}
\begin{CD}
\mathcal{Pic}_{X/k}(S) @>{\iota_S}  >> \Hom(S, \Pic(X/k)) \\
@VV\sigma_s^*V  @VVf_{\sigma_s}V  \\
\mathcal{Pic}_{X/k}(s) @>{id}>> \Hom(s, \Pic(X/k))
\end{CD}
$$
where $\sigma_s^*$ is the pull-back along $\sigma_s$ and $f_{\sigma_s}$ is the pull-back along $\sigma_s$, that is $f_{\sigma_s}(\varphi) = \varphi \circ \sigma_s$ for all $\varphi \in \Hom(S, \Pic(X/k))$. Note that the morphisms $\sigma_s^*$ and $f_{\sigma_s}$ are of this form, by definition of the functors $\mathcal{Pic}(X/k)(\cdot)$ and $\Hom(\cdot, \mathcal{Pic}(X/k))$.
From the commutativity of the diagram, we get that for all line bundle $L$ on $X \times S$:
$$ i_S(L)(s) = f_{\sigma_S}(i_S(L))(s) = \sigma_s^* L = L|_{X \times \{s\}}.$$
If you choose another normalization condition for $i_s$, for instance $i_s(L) = L \otimes L_0$ (for a fixed $L_0$, not depending on $s$), then you would get another homomorphism of functors, say $j$ between $\mathcal{Pic}(X/k)(\cdot)$ and $\Hom(\cdot, \Pic(X/k))$. It would most certainly share the same properties as $\iota$.
