2-commutative diagram in 2-category 
Definition 3.27: A diagram in a 2-category is called 2-commutative, if its 1- morphisms commute up to given 2-isomorphisms and these
2-isomorphisms commute in the induced diagram taking 1-morphisms (and
their compositions) as vertices.

Ref: http://wwwf.imperial.ac.uk/~dg3215/other/stacks.pdf
I have a question about this definition:
Looking only at the simple case of a commutative square, the first part says as follows that the diagram
$$\begin{matrix}
w&\xrightarrow{b}&y\\
\downarrow\rlap{\scriptstyle a}&&\downarrow\rlap{\scriptstyle c}\\
x&\xrightarrow{\scriptstyle d} &z
\end{matrix} $$ 2-commutes if the two 1-morphisms $w \to y \to z$ and $w \to x \to z$ are 2-isomorphic. There, there seems to be one invertible 2-morphism $c \circ b \Rightarrow d \circ a$. (See https://stacks.math.columbia.edu/tag/003O for instance).
Question: What does the second part of the definition mean '2-isomorphisms commute in the induced diagram taking 1-morphisms (and their compositions) as vertices'? Is it possible to illustrate it in a diagram?
UPDATE: Question: Is there a more complicated 'commutative' diagram, like the skeleton of a cube, for which the second part of the definition 3.27 would apply? What is the most commonly used definition of a 2-commutative diagram in a 2-category?
 A: Like is mentioned in the comments, the commutative square in your example doesn't reveal what the second part of the statement is saying simply because there is only one $2$-cell present; this is similar to saying that the diagram
$$
\bullet \longrightarrow\bullet
$$
($1$-)commutes; there's nothing there to check. However, if we pick a slightly bigger diagram, we can start to make better sense of the second part of the statement: for example, to say that
$\require{AMScd}$
\begin{CD}
U @>a>> V @>b>> W \\
@VcVV @VdVV @VVeV \\
X @>>f> Y @>>g> Z
\end{CD}
$2$-commutes, we will need several $2$-cell isomorphisms to satisfy the first part of the definition:

*

*an isomorphism $\alpha:gda\cong gfc$

*an isomorphism $\beta:cba\cong gda$

*an isomorphism $\gamma:cba\cong gfc$
(I'm only considering the various paths $U\to Z$). The second part of the definition of a $2$-commutativity then says that the choice of $2$-cells used here have to be coherent, so composing these isomorphisms recover other isomorphisms present (for example, $\alpha\beta:cba\cong gda\cong gfc$ should not be a different isomorphism than $\gamma:cba\cong gfc$). This can be more concisely stated as $1$-commutativity of a diagram whose vertices are $gda$, $gfc$, and $cba$ (forgive the typesetting, I'm not sure how to make good diagrams on this site) given by
\begin{CD}
cba @>\gamma >> \\
@V\beta VV @VVV \\
gda @>>\alpha> gfc
\end{CD}
This is just a vamped up version of talking about $1$-commutativity: in this case, commutativity of a diagram could have been put simply as

Any path from the start to the end produces the same morphism

moving one level up, we change "equal" to "isomorphic" and then say

Any path (of $1$-cells) from the start to the end is the same morphism up to some chosen $2$-cell isomorphism.
View these $2$-cells as arrows from their ($1$-cell) domain to their codomain and consider the resulting graph. Now, any path from start to end here is the same $2$-cell (up to equality).

These are the two parts of the definition of $2$-commutativity.
This idea that the $2$-cell isomorphisms witnessing commutativity have to be coherent in this way comes from the fact that commutativity of higher diagrams should allow you to somehow "contract" everything down without leaving a mess.
This is similar to the coherence axioms for a monoidal category (since monoidal categories are equivalently one-object bicategories, this is often a helpful intermediate step when trying to gain intuition about $2$-category theory after having a decent grip of $1$-category theory): associativity for example is no longer strict but holds up to $2$-cell isomorphism, and to ensure that we have the desired coherence theorem (which says that a monoidal category is always equivalent to one that is strictly associative and unital), we need the associativity isomorphisms to be coherent with each other, so any two ways of using the associativity isomorphisms to go from e.g. $a\otimes(b\otimes(c\otimes d))$ to $((a\otimes b)\otimes c)\otimes d$ ought to be the same thing (this is the purpose of the pentagon identity).
Edit: To elaborate a bit more on what I meant by "contracting" everything, one way to think about contexts where all higher cells (i.e., $k$-cells for $k\geq2$) are invertible is too look at them as a kind of homotopy theory, in which case everything being uniquely defined "up to homotopy" is another way of saying "up to a contractible space of choices." For example, if you look at a terminal object in an arbitrary $(\infty,1)$-category (for these purposes, it is enough to think of this as a category with $k$-cells for all $k\geq0$, and the $k$-cells are invertible for $k>1$), every time you make a choice, it has to be "unique relative to higher homotopy":

*

*the terminal object is an object $x$. This choice is unique up to higher homotopy (i.e., 1-cell)

*this means, for any other candidate object $y$, there is a $1$-cell $f:y\to x$. This choice of equivalence is unique up to higher homotopy

*this means, for any other candidate equivalence $g:y\to x$, there is a $2$-cell equivalence $\alpha:f\Rightarrow g$. This choice is unique up to higher homotopy

In particular, the subcategory $T$ of all possible terminal objects is equivalent to the trivial category; i.e., $T\simeq*$ is a contractible space of choices of terminal objects.
In the case of $1$-categories, the only $2$-cells are identities (i.e., equality), so $\alpha:f\Rightarrow g$ is actually saying $f=g$ and the universal property of the terminal object reduces to saying for any candidate $y$ that there is a unique arrow $f:y\to x$. For a $2$-category, the $1$-cells need not be unique, but the $2$-cells will be, so $\alpha:f\Rightarrow g$ is unique for any pair $f,g:y\to x$.
This brings us back to $2$-commutative diagrams. If we look at "$\infty$-commutative" diagrams, we're saying that

*

*any two paths of $1$-cells produces the same morphism up to higher homotopy; i.e., up to some path of $2$-cells

*this path of $2$-cells is unique up to higher homotopy, so any two paths of $2$-cells demonstrating this commutativity must be related by some path of $3$-cells

*so on and so forth

In particular, in a $2$-category (so we're talking about $2$-commutativity), the path of $2$-cells witnessing commutativity has to be unique, and this recovers the second constraint in the definition of $2$-commutativity you provided: the diagram of $2$-cells has to be $1$-commutative.
