Lemma for showing that presheaves are colimits of representables $\newcommand{\PShv}{\text{PShv}}$
$\newcommand{\Fun}{\text{Fun}}$
$\newcommand{\C}{\mathcal{C}}$
$\newcommand{\Hom}{\text{Hom}}$
$\newcommand{\ra}{\rightarrow}$
$\newcommand{\op}{\text{op}}$
$\newcommand{\Set}{\mathsf{Set}}$
Let $\C$ be a small category and let $\PShv(\C) := \Fun(\C^\op, \Set)$ be the category of presheaves. I would like to show that for two presheaves $F, G$ we have the natural isomorphism
$$ \Hom_{\PShv(\C)}(F, G) \cong \Hom_{\Fun((* \Rightarrow F)^\op , \PShv(\C))}(P, \Delta(G))  $$
where $(* \Rightarrow F)^\op$ is the (opposite) comma category, $P : (* \Rightarrow F)^\op \ra \PShv(\C)$ is given by $(x, y) \mapsto Y(x)$ ($Y: \C \ra \PShv(\C)$ is the Yoneda embedding $ x \mapsto \Hom_\C(-, x)$) and
$$ \Delta : \PShv(\C) \ra \Fun((* \Rightarrow F)^\op , \PShv(\C))$$
is the diagonal/constant functor (used in the adjoint definition/formulation of (co)limits).
The proof I am reading starts by taking a natural transformation $\eta : F \Rightarrow G \in \Hom_{\PShv(\C)}(F, G)$, and then defines the correspoding natural transformation $\zeta \in \Hom_{\Fun((* \Rightarrow F)^\op , \PShv(\C))}(P, \Delta(G))$ by 
$$ \zeta_{(x, y)}(g) = G(g) (\eta_x(y))  \qquad (*)$$
for any object $(x, y) $ of $(* \Rightarrow F)^\op$ and any $g \in \Hom_\C(x', x)$. So here $\zeta : P \Rightarrow \Delta(G)$ and its component at $(x, y)$ is a map with domain and codomain
$$ \zeta_{(x, y)} : P(x, y) = Y(x) \ra \Delta(G)(x, y). $$
My first question is: how can $\zeta_{(x, y)}$ possibly take a map $g \in \Hom_\C(x', x)$ as its argument if the domain of $\zeta_{(x, y)}$ is $P(x,y)=Y(x)=\Hom_\C(-, x) \in \PShv(\C)$?
The proof then considers the RHS of the isomorphism that is to be proved, namely it states that any $\zeta \in \Hom_{\Fun((* \Rightarrow F)^\op , \PShv(\C))}(P, \Delta(G))$ must be of the form $(*)$ for a unique $\eta \in \Hom_{\PShv(\C)}(F, G)$ by definition of naturality, since $\zeta$ is itself a natural transformation. I'm trying to show this explicitly but I'm not quite sure how to. 
I've started by writing out $\zeta$ as a natural transformation; we have $\zeta : P \Rightarrow \Delta(G)$ natural so for $(x_1, y_1), (x_2, y_2) \in (* \Rightarrow F)^\op$ and a morphism $f: (x_1, y_1) \ra (x_2, y_2)$ we have the naturality square (in equations):
$$ \Delta(G)(f) \circ \zeta_{(x_1, y_1)} = \zeta_{(x_2, y_2)} \circ P(f) $$
I'd be grateful if anybody could explain how to recover something of the form $(*)$ from what we have above. Thank you.
 A: For your first question, it's a matter of sloppiness : what should be written is $(\zeta_{(x,y)})_{x'}(g)$, they just removed the $x'$ for ease of notation. I won't do this in my answer though. 
For you to find $\eta$ you need to unravel the Yoneda lemma (it's no wonder the result you're trying to prove is also sometimes called the Yoneda lemma). I will change your notations slightly, and write $(x,f)$ for a generic element of $\int_CF= (*\implies F)^{op}$, that is $x\in C, f\in F(x)$, so that I can use $y$ for other generic elements of $C$ (and $g$ for a generic element of $F(y)$)
Suppose you have a natural transformation $\zeta : P\to \Delta (G)$. So in particular for $(x,f)\in\int_CF$ you have a map $\zeta_{(x,f)} : Y(x)\to G$. 
Now recall what the Yoneda lemma tells us precisely about the shape of an arrow $\theta : Y(x)\to G$. It tells us that $\theta_z(f) = G(f^{op})(\theta_x(id_x))$ : this is again by the "stupid" square : 
$\require{AMScd} \begin{CD}
\hom(x,x) @>{\theta_x}>> G(x)\\ @V{\hom(f,x)}VV @VV{G(f^{op})}V\\
\hom(z,x) @>>{\theta_z}> G(z)
\end{CD}$
But then, with $g:x'\to x$, $(\zeta_{(x,f)})_{x'}(g)= G(g^{op})((\zeta_{(x,f)})_x(id_x))$
We have our "$\zeta_{(x,f)}(g)$" (removing the $x'$-index, as you did) and our $G(g)$ (I denoted it $G(g^{op})$ for clarity but of course it's the same).
Let's look at what $(\zeta_{(x,f)})_x(id_x)$ does. It is quite clearly an element of $G(x)$, and if you move $f\in F(x)$, it moves with it. Sounds to me like this is a nice way to define $\eta$ : define $\eta_x : F(x)\to G(x)$ by $\eta_x(f) := (\zeta_{(x,f)})_x(id_x)$. 
By what came just before, $(\zeta_{(x,f)})_{x'}(g)= G(g^{op})(\eta_x(f))$, which is what we wanted. We now only have to show that $\eta$ is natural. 
So suppose you have a map $\alpha : x\to y$, and you want to look at the square (I'm removing the $^{op}$'s, hopefully it's still clear)
$\require{AMScd} \begin{CD}
F(y) @>{\eta_y}>> G(y)\\ @V{F(\alpha)}VV @VV{G(\alpha)}V\\
F(x) @>>{\eta_x}> G(x)
\end{CD}$
Start with $g\in F(y)$, and look at $f:= F(\alpha)(g) \in F(x)$. Then $\alpha : (x,f)\to (y,g)$ is a morphism in $\int_CF$, by definition, therefore we have a commutative diagram 
$\require{AMScd} \begin{CD}
Y(x) @>{\zeta_{(x,f)}}>> G\\ @V{P(\alpha)}VV @VV{id}V\\
Y(y) @>>{\zeta_{(y,g)}}> G
\end{CD}$ (I don't know how to do triangles)
But then, if you evaluate this in $x$ and then $id_x$ you get $(\zeta_{(y,g)})_x(\alpha)=(\zeta_{(x,f)})_x(id_x) = \eta_x(f)$.
And now recall again the explicit version of the Yoneda lemma, which tells us precisely that $(\zeta_{(y,g)})_x(\alpha) = G(\alpha)((\zeta_{(y,g)})_y(id_y)) = G(\alpha)(\eta_y(g))$ so that in the end, $\eta_x(f) = G(\alpha)(\eta_y(g))$. But now remember the definition of $f$ : $\eta_x\circ F(\alpha)(g) = G(\alpha)\circ \eta_y(g)$. 
This holds for any $g$, so the diagram commutes, so $\eta$ is natural; and we have found our $\eta$.
It remains to show that this $\eta$ is unique; but this is easy, as $\eta_x(f) = (\zeta_{(x,f)})_x(id_x)$ by the formula applied for $x'=x, g=id_x$ (I'm just realizing now that of course evaluating in $g=id$ would have given us $\eta$ without having to look for it as I did above - but at least it's confirming what I did, and it shows that in these examples there aren't 100 ways to do it)
