I have a family of categories $\{A_i\}_I$, their product category $\Pi A_i$, a category $B$ and a family of functors $F_i:B\rightarrow A_i$. Then there is also a unique functor $F_I:B\rightarrow \Pi A_i$ such that $F_i=\pi_i F_I$.
Now, I have a diagram of functors and natural transformations involving this $F_I$ and I know that the image of this diagram under the projections $\pi_i$ is commutative for all $i\in I$. More concretely, diagrams of the following type commute: $$\require{AMScd} \begin{CD} \pi_i F_I @>{\pi_i \alpha}>> \pi_i HF_I\\ @V{\pi_i \alpha}VV @VV{\pi_i \beta}V \\ \pi_i HF_I @>>{\pi_i \gamma}> \pi_i G F_I. \end{CD}$$ Can I conclude then that commutativity "lifts" to the product category as well, i.e. that the diagram $$\require{AMScd} \begin{CD} F_I @>{\alpha}>> HF_I\\ @V{\alpha}VV @VV{\beta}V \\ HF_I @>>{\pi_i \gamma}> G F_I. \end{CD}$$ also commutes? And why? I know that functors don't in general reflect commutativity of diagrams, but this could maybe come from the universal property of the product. If so, I can't give the precise argument for it.
Suppose that $f, g : A \to \prod_i{B_i}$ and $\pi_i \circ f = \pi_i \circ g$ for all $i$. Then $f = g$. This is thanks to the uniqueness part of the definition of products: since $\pi_i \circ f : A \to B_i$, there is a unique $u : A \to \prod_i{B_i}$ such that $\pi_i \circ u = \pi_i \circ f = \pi_i \circ g$. By uniqueness, $f = u = g$ since both $f$ and $g$ satisfy the needed equation.
So how does that apply here? We have $\pi_i \beta \circ \pi_i \alpha = \pi_i \gamma \circ \pi_i \alpha$, so $\pi_i(\beta \circ \alpha) = \pi_i (\gamma \circ \alpha)$ for all $i$. If we can conclude from this that $\beta \circ \alpha = \gamma \circ \alpha$, then we're done. Equality of natural transformations and the application of functors on them is componentwise, so we just need to prove that if $\pi_i(f) = \pi_i(g)$ for all $i$, then $f = g$ (where $f$ and $g$ are any two morphisms in $\prod_i B_i$).
This is, despite the slight difference, a special case of the fact at the top of this answer. Let $M$ be the category with two objects and one (non-identity) morphism from one object to the other. We can define two functors $F_f$ and $F_g$ from $M$ to $\prod_i B_i$. One is equal to $f$ at the morphism, and the other is equal to $g$. $\pi_i \circ F_f = \pi_i \circ F_g$ because $\pi_i(f) = \pi_i(g)$. By the fact at the top, $F_f = F_g$, so $f = g$.