Assumption (1) implies that $f$ is surjective and flat. Hence the pullback functor is faithful by Lemma \ref{lemma-faithful}. Let $(V, \varphi)$ and $(W, \psi)$ be two descent data relative to $X \to S$. Set $(V', \varphi') = f^*(V, \varphi)$ and $(W', \psi') = f^*(W, \psi)$. Let $\alpha' : V' \to W'$ be a morphism of descent data for $X'$ over $S$. We have to show there exists a morphism $\alpha : V \to W$ of descent data for $X$ over $S$ whose pullback is $\alpha'$. \medskip\noindent Recall that $V'$ is the base change of $V$ by $f$ and that $\varphi'$ is the base change of $\varphi$ by $f \times f$ (see Remark \ref{remark-easier}). By assumption the diagram $$ \xymatrix{ V' \times_S X' \ar[r]_{\varphi'} \ar[d]_{\alpha' \times \text{id}} & X' \times_S V' \ar[d]^{\text{id} \times \alpha'} \\ W' \times_S X' \ar[r]^{\psi'} & X' \times_S W' } $$ commutes. We claim the two compositions $$ \xymatrix{ V' \times_V V' \ar[r]^-{\text{pr}_i} & V' \ar[r]^{\alpha'} & W' \ar[r] & W } , \quad i = 0, 1 $$ are the same. The reader is advised to prove this themselves rather than read the rest of this paragraph. (Please email if you find a nice clean argument.) Let $v_0, v_1$ be points of $V'$ which map to the same point $v \in V$. Let $x_i \in X'$ be the image of $v_i$, and let $x$ be the point of $X$ which is the image of $v$ in $X$. In other words, $v_i = (x_i, v)$ in $V' = X' \times_X V$. Write $\varphi(v, x) = (x, v')$ for some point $v'$ of $V$. This is possible because $\varphi$ is a morphism over $X \times_S X$. Denote $v_i' = (x_i, v')$ which is a point of $V'$. Then a calculation (using the definition of $\varphi'$) shows that $\varphi'(v_i, x_j) = (x_i, v'_j)$. Denote $w_i = \alpha'(v_i)$ and $w'_i = \alpha'(v_i')$. Now we may write $w_i = (x_i, u_i)$ for some point $u_i$ of $W$, and $w_i' = (x_i, u'_i)$ for some point $u_i'$ of $W$. The claim is equivalent to the assertion: $u_0 = u_1$. A formal calculation using the definition of $\psi'$ (see Lemma \ref{lemma-pullback}) shows that the commutativity of the diagram displayed above says that $$ ((x_i, x_j), \psi(u_i, x)) = ((x_i, x_j), (x, u'_j)) $$ as points of $(X' \times_S X') \times_{X \times_S X} (X \times_S W)$ for all $i, j \in \{0, 1\}$. This shows that $\psi(u_0, x) = \psi(u_1, x)$ and hence $u_0 = u_1$ by taking $\psi^{-1}$. This proves the claim because the argument above was formal and we can take scheme points (in other words, we may take $(v_0, v_1) = \text{id}_{V' \times_V V'}$). \medskip\noindent At this point we can use Lemma \ref{lemma-fpqc-universal-effective-epimorphisms}. Namely, $\{V' \to V\}$ is a fpqc covering as the base change of the morphism $f : X' \to X$. Hence, by Lemma \ref{lemma-fpqc-universal-effective-epimorphisms} the morphism $\alpha' : V' \to W' \to W$ factors through a unique morphism $\alpha : V \to W$ whose base change is necessarily $\alpha'$. Finally, we see the diagram $$ \xymatrix{ V \times_S X \ar[r]_{\varphi} \ar[d]_{\alpha \times \text{id}} & X \times_S V \ar[d]^{\text{id} \times \alpha} \\ W \times_S X \ar[r]^{\psi} & X \times_S W } $$ commutes because its base change to $X' \times_S X'$ commutes and the morphism $X' \times_S X' \to X \times_S X$ is surjective and flat (use Lemma \ref{lemma-ff-base-change-faithful}). Hence $\alpha$ is a morphism of descent data $(V, \varphi) \to (W, \psi)$ as desired.