Confusion about "horizontal composition" of natural transformations

I'm having trouble with an exercise from Rotman's Homological Algebra. It has to do with what Wikipedia calls "horizontal composition" of natural transformations. Namely, given $F, G:\mathcal{A}\to\mathcal{B}$ and $F^\prime, G^\prime:\mathcal{B}\to\mathcal{C}$ covariant functors and $\sigma:F\to G$ and $\tau:F^\prime\to G^\prime$ natural transformations, the goal is to show that there is a composite natural transformation $\tau\sigma: F^\prime F\to G^\prime G$.

There should be a "natural choice" for the morphism $(\tau\sigma)_A$ associated to any $A\in\operatorname{obj}\mathcal{A}$, and this is where I'm confused. The problem says to define $$(\tau\sigma)_A=\tau_{FA}\sigma_A : F^\prime F(A)\to G^\prime G(A)$$ but this doesn't make sense to me since $\tau_{FA}\in\operatorname{Hom}(F^\prime F(A),G^\prime F(A))$ and $\sigma_A\in\operatorname{Hom}(FA,GA)$ (right?). It's not clear to me why this is the right composition, or that it even is a composition.

If I had to guess, I would say $(\tau\sigma)_A:=\tau_{GA}F^\prime(\sigma_A)$ because this is the only thing I can come up with that actually is a morphism $F^\prime F(A)\to G^\prime G(A)$.

Can anyone clear up my confusion?

Your guess is correct, but besides $\tau_{GA}\circ F'(\sigma_A)$, there is another natural choice, namely: $$(\tau\sigma)_A=G'(\sigma_A)\circ \tau_{FA}\,,$$ and the point is that they are equal.

To be honest, I have no idea where the $\tau_{FA}\sigma_A$ could come from.. Might be a typo..

Note also that if we define natural transformations as functors $\mathcal A\to\mathcal B^\to$, where $\mathcal B^\to$ denotes the arrow category of $\mathcal B$ whose objects are the arrows of $\mathcal B$ and whose morphisms are commutative squares in $\mathcal B$, then we arrive to

$(\tau\sigma)_A \ =\ $ the common composition of square $\tau(\sigma_A)$.