[proofplan]
We first show that the displayed formula really defines a single morphism in $\mathcal{E}$ by applying the naturality of $\theta$ to the morphism $\eta_X:F(X)\to G(X)$ in $\mathcal{D}$. We then prove that these component morphisms are natural in $X\in\mathcal{C}$. For a morphism $f:X\to Y$ in $\mathcal{C}$, the desired naturality square follows by combining the naturality of $\theta$ at $G(f)$ with the naturality of $\eta$ at $f$ and functoriality of $F'$.
[/proofplan]
[step:Use the naturality of $\theta$ to identify the two component formulas]
Fix an object $X\in\mathcal{C}$. Since $\eta:F\Rightarrow G$ is a natural transformation, its component at $X$ is a morphism
\begin{align*}
\eta_X:F(X)\to G(X)
\end{align*}
in $\mathcal{D}$. Since $\theta:F'\Rightarrow G'$ is natural, applying its naturality square to the morphism $\eta_X:F(X)\to G(X)$ gives
\begin{align*}
G'(\eta_X)\circ \theta_{F(X)}
=
\theta_{G(X)}\circ F'(\eta_X).
\end{align*}
Both sides are morphisms in $\mathcal{E}$ from $F'(F(X))$ to $G'(G(X))$. Therefore the formula
\begin{align*}
(\theta * \eta)_X
:=
\theta_{G(X)}\circ F'(\eta_X)
=
G'(\eta_X)\circ \theta_{F(X)}
\end{align*}
defines a morphism
\begin{align*}
(\theta * \eta)_X:F'(F(X))\to G'(G(X))
\end{align*}
for every object $X\in\mathcal{C}$.
[guided]
Fix an object $X\in\mathcal{C}$. The component $\eta_X$ of the natural transformation $\eta:F\Rightarrow G$ is a morphism in $\mathcal{D}$,
\begin{align*}
\eta_X:F(X)\to G(X).
\end{align*}
The natural transformation $\theta:F'\Rightarrow G'$ has components
\begin{align*}
\theta_A:F'(A)\to G'(A)
\end{align*}
for objects $A\in\mathcal{D}$, and naturality of $\theta$ says that for every morphism $u:A\to B$ in $\mathcal{D}$,
\begin{align*}
G'(u)\circ \theta_A
=
\theta_B\circ F'(u).
\end{align*}
We apply this statement with $A=F(X)$, $B=G(X)$, and $u=\eta_X$. This gives
\begin{align*}
G'(\eta_X)\circ \theta_{F(X)}
=
\theta_{G(X)}\circ F'(\eta_X).
\end{align*}
Thus the two proposed composites from $F'(F(X))$ to $G'(G(X))$ are equal. Hence the notation
\begin{align*}
(\theta * \eta)_X
:=
\theta_{G(X)}\circ F'(\eta_X)
=
G'(\eta_X)\circ \theta_{F(X)}
\end{align*}
is well-defined for each object $X\in\mathcal{C}$.
[/guided]
[/step]
[step:Verify naturality of the constructed components]
Let $f:X\to Y$ be a morphism in $\mathcal{C}$. We must prove that the square for the family $((\theta * \eta)_X)_{X\in\mathcal{C}}$ commutes, namely
\begin{align*}
G'(G(f))\circ (\theta * \eta)_X
=
(\theta * \eta)_Y\circ F'(F(f)).
\end{align*}
Using the definition of $(\theta * \eta)_X$, the naturality of $\theta$ at the morphism $G(f):G(X)\to G(Y)$ in $\mathcal{D}$, functoriality of $F'$, and the naturality of $\eta$ at $f$, we compute:
\begin{align*}
G'(G(f))\circ (\theta * \eta)_X
&=
G'(G(f))\circ \theta_{G(X)}\circ F'(\eta_X) \\
&=
\theta_{G(Y)}\circ F'(G(f))\circ F'(\eta_X) \\
&=
\theta_{G(Y)}\circ F'(G(f)\circ \eta_X) \\
&=
\theta_{G(Y)}\circ F'(\eta_Y\circ F(f)) \\
&=
\theta_{G(Y)}\circ F'(\eta_Y)\circ F'(F(f)) \\
&=
(\theta * \eta)_Y\circ F'(F(f)).
\end{align*}
Thus the family $((\theta * \eta)_X)_{X\in\mathcal{C}}$ is natural, so it defines a natural transformation
\begin{align*}
\theta * \eta:F'\circ F\Rightarrow G'\circ G.
\end{align*}
[guided]
Let $f:X\to Y$ be a morphism in $\mathcal{C}$. To prove that the family of morphisms $((\theta * \eta)_X)_{X\in\mathcal{C}}$ is a natural transformation from $F'\circ F$ to $G'\circ G$, we must prove the commutativity condition
\begin{align*}
G'(G(f))\circ (\theta * \eta)_X
=
(\theta * \eta)_Y\circ F'(F(f)).
\end{align*}
The left-hand side first applies the component of $\theta * \eta$ at $X$ and then moves along the image of $f$ under $G'\circ G$. We expand the definition:
\begin{align*}
G'(G(f))\circ (\theta * \eta)_X
=
G'(G(f))\circ \theta_{G(X)}\circ F'(\eta_X).
\end{align*}
Now apply the naturality of $\theta:F'\Rightarrow G'$ to the morphism
\begin{align*}
G(f):G(X)\to G(Y)
\end{align*}
in $\mathcal{D}$. Naturality gives
\begin{align*}
G'(G(f))\circ \theta_{G(X)}
=
\theta_{G(Y)}\circ F'(G(f)).
\end{align*}
Substituting this into the previous expression yields
\begin{align*}
G'(G(f))\circ (\theta * \eta)_X
=
\theta_{G(Y)}\circ F'(G(f))\circ F'(\eta_X).
\end{align*}
Because $F':\mathcal{D}\to\mathcal{E}$ is a functor, it preserves composition, so
\begin{align*}
F'(G(f))\circ F'(\eta_X)
=
F'(G(f)\circ \eta_X).
\end{align*}
Hence
\begin{align*}
G'(G(f))\circ (\theta * \eta)_X
=
\theta_{G(Y)}\circ F'(G(f)\circ \eta_X).
\end{align*}
Next use the naturality of $\eta:F\Rightarrow G$ at the morphism $f:X\to Y$. This naturality condition is
\begin{align*}
G(f)\circ \eta_X
=
\eta_Y\circ F(f).
\end{align*}
Replacing the composite inside $F'$ gives
\begin{align*}
G'(G(f))\circ (\theta * \eta)_X
=
\theta_{G(Y)}\circ F'(\eta_Y\circ F(f)).
\end{align*}
Again using functoriality of $F'$, we split this composite:
\begin{align*}
F'(\eta_Y\circ F(f))
=
F'(\eta_Y)\circ F'(F(f)).
\end{align*}
Therefore
\begin{align*}
G'(G(f))\circ (\theta * \eta)_X
&=
\theta_{G(Y)}\circ F'(\eta_Y)\circ F'(F(f)) \\
&=
(\theta * \eta)_Y\circ F'(F(f)),
\end{align*}
where the last equality is the definition of $(\theta * \eta)_Y$. This is precisely the naturality square for $\theta * \eta$ at $f$.
[/guided]
[/step]
[step:Conclude that horizontal composition is a natural transformation]
The previous step proves the naturality condition for every morphism $f:X\to Y$ in $\mathcal{C}$. Since each component
\begin{align*}
(\theta * \eta)_X:F'(F(X))\to G'(G(X))
\end{align*}
has the correct source and target, the family $((\theta * \eta)_X)_{X\in\mathcal{C}}$ is a natural transformation
\begin{align*}
\theta * \eta:F'\circ F\Rightarrow G'\circ G.
\end{align*}
This is the horizontal composition of $\theta$ and $\eta$.
[/step]
