Yes, this is an application of descent theory (with a bit of care). Let $F = f^{\ast}(G)$ on $X$, a quasi-coherent $O_X$-module by hypothesis. Then for the maps $p_1, p_2: X \times_Y X \rightrightarrows X$ we have an evident composite isomorphism $$\theta: p_1^{\ast}(F) \simeq (f \circ p_1)^{\ast}(G) = (f \circ p_2)^{\ast}(G) \simeq p_2^{\ast}(F)$$ that satisfies the usual cocycle condition; i.e., $\theta$ is a descent datum. Hence, by fpqc descent for quasi-coherent sheaves, we obtain a quasi-coherent $O_Y$-module $G'$ and an $O_X$-linear isomorphism $\alpha: f^{\ast}(G') \simeq F := f^{\ast}(G)$ respecting the descent data on both sides.

We claim that $\alpha = f^{\ast}(\varphi)$ for a unique $O_Y$-linear isomorphism $\varphi: G' \simeq G$ (in particular giving the quasi-coherence of $G$). Uniqueness is immediate since $G'$ has no non-trivial automorphism inducing the identity after applying $f^{\ast}$ (by the easy faithfulness for $f^{\ast}$ via consideration of stalks over local rings), so we can work Zariski-locally on $Y$ and then Zariski-locally on $X$ to arrange that $Y$ is affine and $X = U_1 \cup \dots \cup U_n$ for affine $U_j$. Thus, $X' := \coprod U_j$ is affine and fpqc over $Y$, and it clearly suffices to treat the situation after replacing $X$ with $X'$ for the purposes of proving existence of $\varphi$ over $Y$. Thus, now we may assume $X$ and $Y$ are affine: $X = {\rm{Spec}}(A)$ and $Y = {\rm{Spec}}(B)$.

Let $f':X \times_Y X \rightarrow Y$ be the natural map, and $F' = {f'}^{\ast}(G)$, so $F'$ is naturally identified with each of $p_1^{\ast}(F)$ and $p_2^{\ast}(F)$ compatibly with $\theta$ (via how $\theta$ is defined). The equality of $f'$ with $f \circ p_1$ and $f \circ p_2$ thereby defines two $O_Y$-linear maps $$f_{\ast}(F) \rightrightarrows f'_{\ast}(F')$$ whose equality is the quasi-coherent $G'$ by design. But $G$ is visibly an $O_Y$-submodule of $f_{\ast}(F)$ (since $f$ is surjective and faithfully flat between local rings) and as such is contained inside the equalizer $G'$, so we have $G \subset G'$. The problem of checking equality thereby reduces to comparing stalks at each $y \in Y$.

If $x \in f^{-1}(y)$ is a point then $O_y \rightarrow O_x$ is faithfully flat, so by fpqc descent for modules we see that $$G_y = \ker(F_x \rightrightarrows F_x \otimes_{O_y} O_x).$$ But the right side coincides with $G'_y$ since $G'$ is a quasi-coherent descent of $F$, so $G_y = G'_y$ via the natural map; i.e., the inclusion $G \subset G'$ is an equality on $y$-stalks.

Yes, this is an application of descent theory (with a bit of care). Let $F = f^{\ast}(G)$ on $X$, a quasi-coherent $O_X$-module by hypothesis. Then for the maps $p_1, p_2: X \times_Y X \rightrightarrows X$ we have an evident composite isomorphism $$\theta: p_1^{\ast}(F) \simeq (f \circ p_1)^{\ast}(G) = (f \circ p_2)^{\ast}(G) \simeq p_2^{\ast}(F)$$ that satisfies the usual cocycle condition; i.e., $\theta$ is a descent datum. Hence, by fpqc descent for quasi-coherent sheaves, we obtain a quasi-coherent $O_Y$-module $G'$ and an $O_X$-linear isomorphism $\alpha: f^{\ast}(G') \simeq F := f^{\ast}(G)$ respecting the descent data on both sides.

Let $f':X \times_Y X \rightarrow Y$ be the natural map, and $F' = {f'}^{\ast}(G)$, so $F'$ is naturally identified with each of $p_1^{\ast}(F)$ and $p_2^{\ast}(F)$ compatibly with $\theta$ (via how $\theta$ is defined). The equality of $f'$ with $f \circ p_1$ and $f \circ p_2$ thereby defines two $O_Y$-linear maps $$f_{\ast}(F) \rightrightarrows f'_{\ast}(F')$$ whose equality is the quasi-coherent $G'$ by design. But $G$ is visibly an $O_Y$-submodule of $f_{\ast}(F)$ (since $f$ is surjective and faithfully flat between local rings) and as such is contained inside the equalizer $G'$, so we have $G \subset G'$. The problem of checking equality thereby reduces to comparing stalks at each $y \in Y$.

If $x \in f^{-1}(y)$ is a point then $O_y \rightarrow O_x$ is faithfully flat, so by fpqc descent for modules we see that $$G_y = \ker(F_x \rightrightarrows F_x \otimes_{O_y} O_x).$$ But the right side coincides with $G'_y$ since $G'$ is a quasi-coherent descent of $F$, so $G_y = G'_y$ via the natural map; i.e., the inclusion $G \subset G'$ is an equality on $y$-stalks.