This answer below, in the corollary, is roughly the same as in my comment.
Let $k$ be an algebraically closed field. Let $X$ be a finitely presented $k$-scheme (later it will be a smooth projective $k$-scheme with nontrivial $\text{Pic}^0_{X/k}$). Let $p:Y\to X$ be an fppf morphism.
Proposition. There exists a morphism $i:Z\to Y$ that is a disjoint union of locally closed immersions such that the composition $p\circ i$ is fppf and quasi-finite.
Proof. This argument is a Bertini hyperplane argument. First of all, up to replacing $Y$ by a disjoint union of open affine subschemes, assume that $Y$ is affine. Denote by $Y_{\text{equi}}\subset Y$ the maximal open subset on which $p$ is (locally) equidimensional, i.e., at every point of every fiber of $p$, the irreducible components of the fiber containing the point all have equal dimensions. The restriction of $p$ to $Y_{\text{equi}}$ is still surjective. Thus, without loss of generality, assume that $p$ is (locally) equidimensional. Then on every connected component $Y_\alpha$ of $Y$, the fiber dimension of $p$ is constant. We construct $Z$ as a disjoint union of schemes $Z_\alpha$ where $i_\alpha:Z_\alpha\to Y_\alpha$ is a disjoint union of locally closed immersions such that $p_\alpha\circ i_\alpha$ is a quasi-finite, flat morphism with image equal to the image of $p_\alpha$.
Thus, without loss of generality, assume that $Y$ is equidimensional of dimension $d$. The proof of the existence of $i:Z\to Y$ as above is by induction on $d$. If $d$ equals $0$, then define $Z$ to equal $Y$. Thus, by way of induction, assume that $d>0$ and assume the result is proved for $d-1$.
Let $e:Y\hookrightarrow X\times_{\text{Spec}\ k} \mathbb{A}^n_k$ be a closed immersion of $X$-schemes. For every closed point $x$ in the image of $p$, the fiber $Y_p=p^{-1}(x)$ has only finitely many associated points. For every linear polynomial $t$ in the coordinate ring $k[\mathbb{A}^n_k]$ that vanishes at none of these associated points, multiplication by $t$ is injective on $\mathcal{O}_{Y_p}$. Denote by $H$ the zero scheme of $t$. By the local flatness criterion, there exists an open subset $U$ of $Y$ containing $Y_p\cap H$ such that $U\cap H$ is flat over $X$. Choosing $t$ so that $Y_p\cap H$ is nonempty, $U\cap H \to X$ has image containing $p$ and it is equidimensional of dimension $d-1$ (by Krull's Hauptidealsatz).
Thus, by the induction hypothesis, there exists a locally closed immersion $i_x:Z_x\to U\cap H$ such that $p\circ i_x$ is quasi-finite and flat with image $V_x$ containing $x$. As we vary $x$, the open images $V_x$ cover $X$. Since we assumed that $X$ is quasi-compact, finitely many of these opens suffice to cover $X$. Define $i:Z\to Y$ to be the disjoint union of these finitely many locally closed immersions $i_x$. Since the composition $p\circ i$ is quasi-finite and flat on each of the finitely many connected components, it is quasi-finite and flat. By construction, the image equals the image of $p$. Thus, the lemma is proved by induction on $d$. QED
Lemma. Let $q:Z\to X$ be a quasi-finite, flat morphism that is surjective. There exists a dense open subscheme $j:V\hookrightarrow X$ and a positive integer $n$ such that the kernel of the pullback homomorphism $q^*:\text{Pic}(X)\to \text{Pic}(Z)$ is contained in the kernel of the homomorphism $\text{Pic}(X)\xrightarrow{j^*}\text{Pic}(V)\xrightarrow{n\cdot -} \text{Pic}(V).$
Proof. For every finitely presented, quasi-finite morphism $q:Z\to X$, there exists a dense open subset $V\subset X$ such that $q^{-1}(V)\to V$ is finite. There exists a norm map, $$\text{Nm}_q : \text{Pic}(q^{-1}(V))\to \text{Pic}(V), \ \ \mathcal{L} \mapsto \text{det}(q_*\mathcal{L})\otimes \text{det}(q_*\mathcal{O})^{\vee}.$$ This is described in detail in Mumford's Lectures on curves on an algebraic surface. For every connected component $V_\alpha$ of $V$, the morphism $q$ has constant finite degree $m$ over that component. For every invertible sheaf $\mathcal{M}_\alpha$ on $V_\alpha$, $\text{Nm}_q(q^*\mathcal{M}_\alpha)$ equals $\mathcal{M}_\alpha^{\otimes m}$. Thus, defining $n$ to equal the least common multiple of these integers $m$, for every invertible sheaf $\mathcal{M}$ on $V$, if $q^*\mathcal{M}$ is trivial, then also $\mathcal{M}^{\otimes n}$ is trivial. Thus, for an invertible sheaf $\mathcal{M}$ on $X$, if $q^*\mathcal{M}$ is trivial, then also $j^*\mathcal{M}^{\otimes n}$ is trivial. QED
Corollary. For every smooth, projective $k$-scheme $X$ such that $\text{Pic}^0_{X/k}$ has positive dimension, for every fppf morphism $p:Y\to X$, the image of $\text{Pic}^0_{X/k}(k)$ in $\text{Pic}(Y)$ is an infinitely generated, divisible group.
Proof. The group of $k$-points $\text{Pic}^0_{X/k}(k)$ is divisible, has finite $n$-torsion for every integer $n$, and is infinitely generated. The proposition precisely says that the image of $p^*$ is still infinitely generated.
By the proposition, there exists a morphism $i:Z\to Y$ such that $q=p\circ i:Z\to X$ is fppf and quasi-finite. Since the image of $p^*$ is intermediate between $\text{Pic}^0_{X/k}(k)$ and its image under $q^*$, it suffices to prove that the image under $q^*$ is infinitely generated. By the lemma, there exists a dense open subset $j:V\hookrightarrow X$ and an integer $n$ such that the kernel of $q^*$ is contained in the kernel of $$\text{Pic}^0_{X/k}(k)\xrightarrow{j^*}\text{Pic}(V)\xrightarrow{n\cdot -}\text{Pic}(V).$$
The kernel of the restriction map $j^*:\text{Pic}(X)\to \text{Pic}(V)$ is generated by the finitely many irreducible components of $X\setminus V$ that have codimension $1$ in $X$. Thus, the intersection of this kernel with the subgroup $\text{Pic}^0_{X/k}(k)$ is a subgroup of a finitely generated group, hence it is also finitely generated. The quotient of $\text{Pic}^0_{X/k}(k)$ by this finitely generated subgroup is still divisible and it is still infinitely generated. Moreover, it still has finite $n$-torsion for every integer $n$: the inverse image in $\text{Pic}^0_{X/k}(k)$ of this $n$-torsion is an extension of a subgroup of the finitely generated kernel by the $n$-torsion of $\text{Pic}^0_{X/k}(k)$. Thus the image is an $n$-torsion group that is finitely generated (the image of a finitely generated group), hence it is finite. So, for each integer $n$, the quotient of this group by its $n$-torsion is still an infinitely generated, divisible group. QED