How do you derive the adjoint's "naturality" condition as seen in MacLane & Moerdijk's book "Sheaves in Geometry and Logic"? It's tag (7) as pictured below.  I also included the definition of adjoint that they use.

I know that by definition of adjunction (using the natural homset isomorphism), we have two naturality squares, i.e. for all $\xi : X' \to X$ in our category $C$ we have:
Naturality square in $X$
and similarly a naturality square in $A$, for all $\alpha: A \to A'$ in our category $C$.
However, I'm still having trouble deriving (7) as pictured.  Namely the bijective correspondence between (compositions of) maps:
$X' \xrightarrow{\xi} X \xrightarrow{f} GA \xrightarrow{G\alpha} GA'$ and
$FX' \xrightarrow{F\xi} FX \xrightarrow{h} A \xrightarrow{\alpha} A'$.  And I don't know how to state that in $\LaTeX$ with the horizontal bar, so I just stated it in English.
It seems like they just stated something non-obvious, without any kind of proof, at least to me.  I hate it when books do that.  If you're going to author a book with proofs, then prove things.  Ask an undergrad if it's proved to them and so on...  Other than that, Sheaves in Geometry & Logic is the best beginner book that introduces Sheaves & Topoi in my oppinion.
 A: $\DeclareMathOperator{Hom}{Hom}$There are two functors $B, C : \mathbf{X}^{op} \times \mathbf{A} \to Set$ we are dealing with, defined on objects by $B(X, A) = \Hom_\mathbf{X}(F(X), A)$ and $C(X, A) = \Hom_\mathbf{A}(X, G(A))$. We are discussing what it means for a family of morphisms $\{\theta_{A, X} : B(X, A) \to C(X, A)\}_{(X, A) \in \mathbf{X}^{op} \times \mathbf{A}}$ to be natural.
This is a special case of something more general.
Consider arbitrary categories $\mathbf{W}, \mathbf{Y}, \mathbf{Z}$. Consider functors $G, H : \mathbf{W} \times \mathbf{Y} \to \mathbf{Z}$. What does it mean to have a natural transformation $\theta : G \to H$?
By the definition of naturality, it means that we have, for each object $K \in \mathbf{W} \times \mathbf{Y}$, a chosen map $\theta_K : G(K) \to H(K)$, where the collection $\theta$ of these maps satisfies a naturality condition.
Since, by the definition of the category $\mathbf{W} \times \mathbf{Y}$, every object $K \in \mathbf{W} \times \mathbf{Y}$ is of the form $(W, Y)$ for some $W \in \mathbf{W}$, $Y \in \mathbf{Y}$, this is equivalent to saying that we have, for each such $W$ and $Y$, a chosen map $\theta_{(W, Y)}$ (which we will also write as $\theta_{W, Y}$, dropping the parentheses).
The naturality condition states that for all objects $K, K' \in \mathbf{W} \times \mathbf{Y}$ and all morphisms $f : K \to K'$, the naturality square commutes; that is, $H(f) \circ \theta_K = \theta_{K'} \circ G(f)$.
Again by the definition of the category $\mathbf{W} \times \mathbf{Y}$, we can write each such $K, K'$ as $(W, X)$, $(W', X')$. Furthermore, we can write any such $f : (W, X) \to (W', X')$ uniquely as $(k, j)$ for some $k : W \to W'$, $j : X \to X'$.
The naturality condition therefore states the following:

Condition A (which is also the definition of naturality, applied to this context): A collection of maps $\{\theta_{W, Y} : G(W, Y) \to H(W, Y)\}_{W \in \mathbf{W}, Y \in \mathbf{Y}}$ is a natural transformation $\theta : G \to H$ if, and only if, for all objects $W, W' \in \mathbf{W}$ and $Y, Y' \in \mathbf{Y}$, and for all arrows $k : W \to W'$, $j : Y \to Y'$, the equation $H(k, j) \circ \theta_{W, Y} = \theta_{W', Y'} \circ G(k, j)$ holds.

Taking a brief break from this general condition, let's go back to the special case of $B$ and $C$ from our adjunction. If you analyse the definition of naturality presented in Sheaves, you will find that it is exactly the above condition.
Let's return to the general case. One can also imagine picking some other definition for naturality.
In particular, fix some object $W \in \mathbf{W}$. Then we can partially apply the functors $G, H$ to $W$ to get functors $G(W, -), H(W, -) : \mathbf{Y} \to \mathbf{Z}$ - call these functors $G' = G(W, -)$ and $H' = H(W, -)$. We can also partially apply $\theta$ to get a family of maps $\{\theta_{W, Y} : G'(Y) \to H'(Y)\}_{Y \in \mathbf{Y}}$ - call this partial application $\theta'_{-} = \theta_{W, -}$. We could then discuss what it would mean for $\theta'$ to be a natural transformation $\theta' : G' \to H'$.
Similarly, we could do exactly the same thing by partially applying our functors to some $Y \in \mathbf{Y}$.
We could thus arrive at a different condition for naturality:

Condition B: A family $\{\theta_{W, Y} : G(W, Y) \to H(W, Y)\}_{W \in \mathbf{W}, Y \in \mathbf{Y}}$ satisfies Condition B if and only if all the $\theta'$s mentioned above are natural transformations (and similarly if we fix the $Y$ instead of the $W$).

Condition B is the "Two Commutative Squares" definition that you alluded to in your question. By contrast, Condition A is a "One Commutative Square" definition.
We can rephrase Condition B more precisely as

Condition B: $\theta$ satisfies condition $B$ if and only if (1) for all $W \in \mathbf{W}$ and $Y, Y' \in \mathbf{Y}$, and for all $j : Y \to Y'$, we have $H(1_W, j) \circ \theta_{W, Y} = \theta_{W, Y'} \circ G(1_W, j)$, and (2) for all $W, W' \in \mathbf{W}$ and $Y \in \mathbf{Y}$, and for all $k : W \to W'$, we have $H(k, 1_Y) \circ \theta_{W, Y} = \theta_{W', Y} \circ G(k, 1_W)$.

It turns out:

Thm. $\theta$ is a natural transformation $\theta : G \to H$ if and only if it satisfies Condition B.

It's easy to show that any natural $\theta$ satisfies condition B. Both parts (1) and (2) are just special cases of the definition of naturality.
It's a bit more work to show that if $\theta$ satisfies condition $B$, then it also satisfies condition $A$. But you can express the relevant diagram you need to show commutative in condition $A$ as the pasting of two different condition $B$ diagrams (in fact, you can do so in 2 different ways). In particular, we have
$\begin{equation}
\begin{split}
H(k, j) \circ \theta_{W, Y} 
&= H((k, 1_{Y'}) \circ (1_W, j)) \circ \theta_{W, Y} \\
&= H(k, 1_{Y'}) \circ H(1_W, j) \circ \theta_{W, Y} \\
&= H(k, 1_{Y'}) \circ \theta_{W, Y'} \circ G(1_W, j) \\
&= \theta_{W', Y'} \circ G(k, 1_{Y'}) \circ G(1_W, j) \\
&= \theta_{W', Y'} \circ G((k, 1_{Y'}) \circ (1_W, j)) \\
&= \theta_{W', Y'} \circ G(k, j)
\end{split}
\end{equation}$
Of course, the other way to do it is to split $(k, j) = (1_{W'}, j) \circ (k, 1_Y)$.
This shows that the definition of naturality in Sheaves is equivalent to your definition.
One last note. We can generalise this to any finite product - eg, we can do the same thing for characterising natural transformations between functors $G, H : \mathbf{W}_1 \times \cdots \times \mathbf{W}_n \to \mathbf{Z}$, showing that such a transformation is natural if and only if it makes $n$ different sorts of diagrams commute where all components but 1 are fixed. But we cannot generalise this to infinite products.
A: I was able to answer my question, but instead of deleting the OP, I thought I'd make an answer since I know at least one other person reading this book, namely @Shaun.
Basically, you take two creatively chosen naturality squares in $\textbf{Set}$ or wherever the homsets happen to reside, which hold for all $X,X', A, A'$ in their respective categories $\textbf{X}, \textbf{A}$, and you glue them together!
See this commutative diagram.  You then read off the outter rectangle which commutes by the homset isomorphism defining adjointness of $F$ with $G$, where $\theta$ is the natural isomorphism between $\text{Hom}(-, G-)$ and $\text{Hom}(F-, -)$ (bifunctors).  What you get is:
$$
\theta(G\alpha \circ f \circ \xi) = \theta \circ G\alpha \circ (-\circ \xi) \circ f = \alpha \circ (-\circ F\xi) \circ \theta 
\circ f = \alpha \circ \theta(f) \circ F\xi \tag{9}
$$
which is exactly what their notation reads.  Theta is the bijection relating those two compositions.   To go the other way, you could just substitute in identity morphisms wherever appropriate and you get back the two naturality squares in the standard definition of adjunction.
I've tagged the above with (9), which is the tag of the same equation in the book (right after the OP screenshot).
$\blacksquare$
