Let $S,S'$ be schemes, let $\pi : S' \to S$ be a morphism which is faithfully flat and locally of finite presentation, set $S'' := S' \times\_{S} S'$ and $S''' := S' \times\_{S} S' \times\_{S} S'$ with projections $p\_{1},p\_{2} : S'' \to S'$ and $p\_{12},p\_{13},p\_{23} : S''' \to S''$. Let $\mathcal{E}'$ be a vector bundle on $S'$ such that there is an isomorphism $\varphi : p\_{1}^{\ast}\mathcal{E}' \to p\_{2}^{\ast}\mathcal{E}'$ of $\mathcal{O}\_{S''}$-modules. What's an example of such $S,S',\pi,\mathcal{E}'$ such that $\mathcal{E}'$ does not descend to $S$, i.e. there does not exist a vector bundle $\mathcal{E}$ on $S$ and an $\mathcal{O}\_{S'}$-module isomorphism $\pi^{\ast}\mathcal{E} \simeq \mathcal{E}'$?


*Thoughts*:


* Such $\mathcal{E}$ exists if and only if there exists some $\psi \in \operatorname{Aut}\_{\mathcal{O}\_{S''}}(p\_{1}^{\ast}\mathcal{E}')$ such that the cocycle condition $p\_{23}^{\ast}(\varphi \circ \psi) \circ p\_{12}^{\ast}(\varphi \circ \psi) = p\_{13}^{\ast}(\varphi \circ \psi)$ is satisfied.
* If $\pi$ admits a section $\sigma : S \to S'$, then we can take $\mathcal{E} = \sigma^{\ast}\mathcal{E}'$ and pull back $\varphi$ via an induced section $S' \to S''$.
* If $S = \operatorname{Spec} k$ and $S'$ is a smooth projective geometrically integral $k$-scheme and $\mathcal{E}'$ is a line bundle, then $\operatorname{Pic}(S') \times \operatorname{Pic}(S') \to \operatorname{Pic}(S'')$ is injective so the existence of $\varphi$ implies that $\mathcal{E}' \simeq \mathcal{O}\_{S'}$.



A noneffective descent datum: isomorphism not satisfying the cocycle condition
This already fails for line bundles on smooth projective curves: let $X$ be 'the' pointless conic over $\mathbb{R}$, given by the closed subscheme of $\mathbb{P}^2\_{\mathbb{R}}$ cut out by $X^2+Y^2+Z^2 = 0$.
It is smooth, projective, geometrically integral over $\mathbb{R}$, and $X(\mathbb{R})= \emptyset$.
In your notation, we let $\pi\colon S' \rightarrow S$ be the fppf morphism $X\_{\mathbb{C}} \rightarrow X$. In this case fppf descent translates into Galois descent.


Since $X\_{\mathbb{C}} \simeq \mathbb{P}^1\_{\mathbb{C}}$, the isomorphism class of a line bundle on $X\_{\mathbb{C}}$ is determined by its degree.
Since the action of $Gal(\mathbb{C}|\mathbb{R})$ preserves degrees of divisors, we have that $L \simeq c^\*L$ if $c$ denotes complex conjugation $X\_{\mathbb{C}} \rightarrow X\_{\mathbb{C}}$.
Therefore a line bundle $L$ on $X\_{\mathbb{C}}$ of degree $1$ gives an example you're looking for.
Indeed we have just seen that $c^\* L\simeq L$, and if there were a line bundle $M$ on $X\_{\mathbb{R}}$ complexifying to $L$, it would need to have degree $1$ hence (by Riemann-Roch) be effective, which is impossible since $X(\mathbb{R})=\emptyset$.


