My textbook:
The naturality axiom implies that from each array of maps $A_0 \rightarrow... \rightarrow A_n$, $F(A_n) \rightarrow B_0$, $B_0 \rightarrow... \rightarrow B_m$ it is possible to construct exactly one map $A_0 \rightarrow G(B_m)$.
$F, G$ are adjoint functors.
The naturality axiom has two parts: $$\overline{F(A) \rightarrow B \rightarrow B'} = (A \rightarrow G(B) \rightarrow G(B')$$ $$\overline{A' \rightarrow A \rightarrow G(B)} = (F(A') \rightarrow F(A) \rightarrow B)$$
Is it because both left part ($A_0 \rightarrow... \rightarrow A_n$) and right part ($B_0 \rightarrow... \rightarrow B_m$) are trivially reduced to $f = A_0 \rightarrow A_n$ and $g = B_0 \rightarrow B_m$ respectively and thus there is only one way to combine $g \circ F(A_n) \rightarrow B_0 \circ f$?
Suppose $F ⊢ G : → ℬ$ with $$fᵢ : Aᵢ → Aᵢ₊₁ ∈ $$ and $$gᵢ : Bᵢ → Bᵢ₊₁ ∈ ℬ$$ and $$m : F Aₙ → B₀$$
Then we want to construct a map $A₀ → G Bₘ$.
Indeed the transpose of $m$ yields $$m′ : Aₙ → G B₀$$ and the $G$-lifting of the $gᵢ$ yields $$G gᵢ : G Bᵢ → G Bᵢ₊₁$$
Hence, $$G gₘ₋₁ ∘ G gₘ₋₂ ∘ G g₀ ∘ ⋯ ∘ m′ : Aₙ → G Bₘ$$ Whence, $$G gₘ₋₁ ∘ G gₘ₋₂ ∘ G g₀ ∘ ⋯ ∘ m′ ∘ fₙ₋₁ ∘ ⋯ f₀ : A₀ → G Bₘ$$
There might be other maps $A₀ → G Bₘ$, but from the given data I am only able to construct the one above.