Here is the definition of naturality from pg.50 (Problem 2.1.14) in Tom Leinster "Basic Category Theory":
$\overline {A' \xrightarrow {\text{p}} A \xrightarrow {\text{f}} G(B) \xrightarrow {\text{G(q)}} G(B')} = (F(A') \xrightarrow {\text{F(p)}} F(A) \xrightarrow{\bar{f}} B \xrightarrow {\text{q}} B')$
For all $p,f$ and $q.$
And here is the definition of commutativity of a diagram in general which in algebraic topology is usually interpreted as naturality
Here is an example of a commutative diagram:
$$![commutative triangle][1]$$
We have the above triangle such that $$g\circ f = h.$$
Here's another example:
Where we have the above square such that $$h\circ f = k\circ g.$$
Could someone tell me how can we go from the general definition of naturality given by Tom Leinster to the definition of naturality via the commutative square given above?
If you follow Leinster carefully (precisely, Chapter 4, after he introduces Representability and the Hom bifunctor), he gives the precise definition of what he means by the "naturality" assumption in the definition of an adjunction. Namely, given a pair of functors in opposite directions : $$F : \mathcal{A} \to \mathcal{B},\text{ }G : \mathcal{B} \to \mathcal{A}$$ an adjunction between $F$ and $G$ (with $F \dashv G$) is a natural isomorphism $$\theta : \mathcal{A}(-,G(-)) \xrightarrow{\sim} \mathcal{B}(F(-),-)$$, both of which are bifunctors $\mathcal{A}^{op} \times \mathcal{B} \to Set$
If you unravel the definition of a natural transformation (as defined in Chapter 1), you will get the required relation in the definition of an adjunction (which you have mentioned in the beginning).
Precisely, the data of a natural isomorphism $\theta$ as above consists of precisely, for each object $(A,B) \in \mathcal{A}^{op} \times \mathcal{B}$, an isomorphism $\theta_{(A,B)} : \mathcal{A}(A,GB) \xrightarrow{\sim} \mathcal{B}(FA,B)$, mapping $f \mapsto \overline {f}$, such that for any arrow $(A,B) \xrightarrow{(p^{op},q)} (A',B')$ in the product category $\mathcal{A}^{op} \times \mathcal{B}$, the following square commutes in Set :
$\require{AMScd}$ \begin{CD} \mathcal{A}(A,GB) @>{\theta_{(A,B)}}>> \mathcal{B}(FA,B)\\ @V{G(q) \circ - \circ p}VV @V{q \circ - \circ F(p)}VV\\ \mathcal{A}(A',GB') @>{\theta_{(A',B')}}>> \mathcal{B}(FA',B') \end{CD}
Take any map $f : A \to GB$ in $\mathcal{A}$. Following the path (top arrow, right arrow) in the above square gives you the map $q \circ \overline{f} \circ F(p)$. Following the path (left arrow, bottom arrow) in the above square gives you the map $\overline{G(q) \circ f \circ p}$. The commutative of the above square suggests the equality of these two maps.
Conversely, the existence of a bijection $\mathcal{A}(A,GB) \xrightarrow{\sim} \mathcal{B}(FA,B)$ for each $A \in \mathcal{A}$, $B \in \mathcal{B}$ subject to the above equality (as given in the definition of an adjunction by Leinster in Chapter 2) will give you a natural isomorphism $\mathcal{A}(-,G(-)) \xrightarrow{\sim} \mathcal{B}(F(-),B)$.