Functor $F$ is equivalence of categories implies $F$ is full. I am trying to understand the answer given to this question.
From what I understand they are saying that if there exists an $f$ such that $Ff = g$ then by naturality it must have the property that $GFf = Gg$ from which faithfulness of $G$ implies $Ff = g$.
The problem is that I don't see why we are guaranteed to have a morphism $f$ with this property. The definition of a natural transformation states that for every $f: X \rightarrow Y$ we have $\eta_Y \circ F(f) = G(f) \circ \eta_X$, so I do not see how we can use naturality unless we already know some $f$ exists with $Ff = g$.
 A: You are misreading the answer.  They aren't assuming such an $f$ exists: they are explaining that you can find the formula for such an $f$ by assuming it exists and figuring out what it must be.  Namely if $f:X\to Y$ exists, it must be given by the formula $f=\eta_Y^{-1} \circ Gg \circ \eta_X$.
Now stop assuming that $f$ satisfies $Ff=g$ and simply define $f=\eta_Y^{-1} \circ Gg \circ \eta_X$.  You can now prove directly from this formula (using naturality of $\eta$ several times) that $GFf=Gg$, and so $Ff=g$.
Here are the details of the proof that $GFf=Gg$.  We have $$GFf=GF\eta_Y^{-1}\circ GFGg\circ GF\eta_X.$$  Now by naturality of $\eta$ (applied to the map $\eta_X:X\to GFX$), $GF\eta_X\circ\eta_X=\eta_{GFX}\circ\eta_X$.  Since $\eta_X$ is an isomorphism, $GF\eta_X=\eta_{GFX}$.  Similarly, $GF\eta_Y=\eta_{GFY}$, so we have $$GFf=\eta_{GFY}^{-1}\circ GFGg\circ \eta_{GFX}.$$
Finally, naturality of $\eta$ (applied to the map $Gg:GFX\to GFY$) gives $GFGg\circ\eta_{GFX}=\eta_{GFY}\circ Gg$, and so $$GFf=Gg.$$
