[proofplan] We define the proposed composite natural transformation objectwise by composing the component of $\eta$ with the component of $\theta$. The only point to verify is naturality: for each morphism $f:X\to Y$ in $\mathcal{C}$, the two composites from $F(X)$ to $H(Y)$ must agree. This follows by first using naturality of $\theta$, then naturality of $\eta$, together with associativity of composition in $\mathcal{D}$. [/proofplan] [step:Check that each component has the required source and target] Let $X\in\operatorname{Ob}(\mathcal{C})$ be an object. 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 $\theta:G\Rightarrow H$ is a natural transformation, its component at $X$ is a morphism \begin{align*} \theta_X:G(X)\to H(X) \end{align*} in $\mathcal{D}$. Therefore the composite \begin{align*} \theta_X\circ\eta_X:F(X)\to H(X) \end{align*} is defined in $\mathcal{D}$. Thus the formula \begin{align*} (\theta\circ\eta)_X:=\theta_X\circ\eta_X \end{align*} assigns to each object $X$ of $\mathcal{C}$ a morphism from $F(X)$ to $H(X)$. [/step] [step:Verify the naturality square for an arbitrary morphism] Let $f:X\to Y$ be a morphism in $\mathcal{C}$. We must prove that the following equality of morphisms $F(X)\to H(Y)$ in $\mathcal{D}$ holds: \begin{align*} H(f)\circ(\theta\circ\eta)_X=(\theta\circ\eta)_Y\circ F(f). \end{align*} Using the definition of $(\theta\circ\eta)_X$ and associativity of composition in $\mathcal{D}$, we compute \begin{align*} H(f)\circ(\theta\circ\eta)_X &=H(f)\circ(\theta_X\circ\eta_X)\\ &=(H(f)\circ\theta_X)\circ\eta_X. \end{align*} By naturality of $\theta:G\Rightarrow H$ applied to the morphism $f:X\to Y$, we have \begin{align*} H(f)\circ\theta_X=\theta_Y\circ G(f). \end{align*} Substituting this equality gives \begin{align*} (H(f)\circ\theta_X)\circ\eta_X &=(\theta_Y\circ G(f))\circ\eta_X\\ &=\theta_Y\circ(G(f)\circ\eta_X), \end{align*} again by associativity of composition in $\mathcal{D}$. By naturality of $\eta:F\Rightarrow G$ applied to $f:X\to Y$, we have \begin{align*} G(f)\circ\eta_X=\eta_Y\circ F(f). \end{align*} Substituting this equality and reassociating gives \begin{align*} \theta_Y\circ(G(f)\circ\eta_X) &=\theta_Y\circ(\eta_Y\circ F(f))\\ &=(\theta_Y\circ\eta_Y)\circ F(f)\\ &=(\theta\circ\eta)_Y\circ F(f). \end{align*} Hence \begin{align*} H(f)\circ(\theta\circ\eta)_X=(\theta\circ\eta)_Y\circ F(f), \end{align*} which is the naturality condition for the family $(\theta\circ\eta)_X$. [guided] Fix a morphism $f:X\to Y$ in $\mathcal{C}$. The naturality condition for the proposed transformation $\theta\circ\eta:F\Rightarrow H$ asks that the two ways of going from $F(X)$ to $H(Y)$ agree: \begin{align*} H(f)\circ(\theta\circ\eta)_X=(\theta\circ\eta)_Y\circ F(f). \end{align*} We start with the left-hand side and expand the definition of the component $(\theta\circ\eta)_X$: \begin{align*} H(f)\circ(\theta\circ\eta)_X &=H(f)\circ(\theta_X\circ\eta_X)\\ &=(H(f)\circ\theta_X)\circ\eta_X. \end{align*} The second equality is associativity of composition in the category $\mathcal{D}$. Now use the naturality of $\theta:G\Rightarrow H$. Applied to the morphism $f:X\to Y$, naturality says that the square comparing $G(f)$ and $H(f)$ commutes: \begin{align*} H(f)\circ\theta_X=\theta_Y\circ G(f). \end{align*} Substituting this into the previous expression gives \begin{align*} (H(f)\circ\theta_X)\circ\eta_X &=(\theta_Y\circ G(f))\circ\eta_X\\ &=\theta_Y\circ(G(f)\circ\eta_X), \end{align*} where the last equality is again associativity of composition in $\mathcal{D}$. Now use the naturality of $\eta:F\Rightarrow G$. Applied to the same morphism $f:X\to Y$, naturality gives \begin{align*} G(f)\circ\eta_X=\eta_Y\circ F(f). \end{align*} Substituting this equality yields \begin{align*} \theta_Y\circ(G(f)\circ\eta_X) &=\theta_Y\circ(\eta_Y\circ F(f))\\ &=(\theta_Y\circ\eta_Y)\circ F(f)\\ &=(\theta\circ\eta)_Y\circ F(f). \end{align*} Thus the two composites from $F(X)$ to $H(Y)$ agree: \begin{align*} H(f)\circ(\theta\circ\eta)_X=(\theta\circ\eta)_Y\circ F(f). \end{align*} This is exactly the naturality condition for the family $(\theta\circ\eta)_X$. [/guided] [/step] [step:Conclude that the componentwise composite is a natural transformation] The previous steps show that for every object $X$ of $\mathcal{C}$, the morphism $(\theta\circ\eta)_X:F(X)\to H(X)$ is defined in $\mathcal{D}$, and for every morphism $f:X\to Y$ in $\mathcal{C}$ the naturality equation \begin{align*} H(f)\circ(\theta\circ\eta)_X=(\theta\circ\eta)_Y\circ F(f) \end{align*} holds. Therefore the family $((\theta\circ\eta)_X)_{X\in\operatorname{Ob}(\mathcal{C})}$ is a natural transformation $F\Rightarrow H$. [/step]