The way I understand that equation is in terms of bijective correspondences.

An adjunction, crudely, says that arrows \\(F(a)\rightarrow b\\) in \\(\mathcal{B}\\) correspond precisely to arrows \\(a\rightarrow G(b)\\) in \\(\mathcal{A}\\) .

Let's write \\(h^*\\) for the \\(\mathcal{A}\\)-arrow corresponding to \\(h : F(a)\rightarrow b\\) in \\(\mathcal{B}\\).

(This is just a more compact way of writing the effect of \\(\alpha\\), ie \\(h \mapsto h^*\\) is the component \\(\alpha_{a,b}\\).)

But actually adjunction is a bit more than that – the correspondence has to be _natural_, ie it plays nicely with composition in both categories.

So if we post-compose \\(h\\) with some \\(g : b\rightarrow b'\\) in \\(\mathcal{B}\\), we want \\(g\circ h\\) to correspond to \\(G(g)\circ h^*\\).

Similarly, if we pre-compose \\(h^* \\) with some \\(f : a'\rightarrow a\\) in \\(\mathcal{A}\\), we want \\(h^*\circ f\\) to be the correspondent of \\(h\circ F(f)\\).

Putting these together we get \\((g\circ h\circ F(f))^* = G(g)\circ h^*\circ f\\), which is what the naturality square says.

An adjunction, crudely, says that arrows \\(F(a)\rightarrow b\\) in \\(\mathcal{B}\\) correspond precisely to arrows \\(a\rightarrow G(b)\\) in \\(\mathcal{A}\\) .

Let's write \\(h^*\\) for the \\(\mathcal{A}\\)-arrow corresponding to \\(h : F(a)\rightarrow b\\) in \\(\mathcal{B}\\).

(This is just a more compact way of writing the effect of \\(\alpha\\), ie \\(h \mapsto h^*\\) is the component \\(\alpha_{a,b}\\).)

But actually adjunction is a bit more than that – the correspondence has to be _natural_, ie it plays nicely with composition in both categories.

So if we post-compose \\(h\\) with some \\(g : b\rightarrow b'\\) in \\(\mathcal{B}\\), we want \\(g\circ h\\) to correspond to \\(G(g)\circ h^*\\).

Similarly, if we pre-compose \\(h^* \\) with some \\(f : a'\rightarrow a\\) in \\(\mathcal{A}\\), we want \\(h^*\circ f\\) to be the correspondent of \\(h\circ F(f)\\).

Putting these together we get \\((g\circ h\circ F(f))^* = G(g)\circ h^*\circ f\\), which is what the naturality square says.