Having trouble proving natural transformation horizontal composition equality of two formulas using a diagram.

Let $C, C', C''$ be three categories. Let $C \xrightarrow{F_1, F_2} C'$ and $C' \xrightarrow{G_1, G_2} C''$, be four functors, and let $\theta : F_1 \Rightarrow F_2, \ \lambda : G_1 \Rightarrow G_2$, be two natural transformations. Then I've already shown algebraically that if we defined $(\lambda * \theta)_X \equiv G_2(\theta_X) \circ \lambda_{F_1 X}$, then $\lambda * \theta$ is a natural transformation $G_1 \circ F_1 \Rightarrow G_2 \circ F_2$. Now I want to show that for all $X \in C$, the following diagram commutes (or the two formulas for $\lambda * \theta$ are the same. A book says by naturality of $\theta$ and functoriality of $G_1$ the square commutes. But I'm having trouble seeing this. Please explain using diagrams or give a hint.

You have already written the proof of this fact.

Definition. Let $\mathcal{A}$ and $\mathcal{B}$ be categories, $T,S\colon\mathcal{A}\to\mathcal{B}$ be functors, $\alpha\colon\text{Obj}(\mathcal{A})\to\text{Mor}(\mathcal{B})$ be a function. Then the function $\alpha$ is called a natural transformation iff for every morphism $f\colon a_1\to a_2$ in $\mathcal{A}$ the following equality holds: $$ S(f)\circ\alpha(a_1)=\alpha(a_2)\circ T(f). $$

Now assign $\mathcal{A}=C'$, $\mathcal{B}=C''$, $T=G_1$, $S=G_2$, $\alpha=\lambda$, $a_1=F_1(x)$, $a_2=F_2(x)$, $f=\theta(x)$ and we get: $$ G_2(\theta(x))\circ\lambda(F_1(X))=\lambda(F_2(x))\circ G_1(\theta(x)), $$ as required. The commutativity of your diagram is nothing but this equality.

Here's a more diagramatically-based proof.