I've always thought that "canonical" means "unique..." having a certain (universal, sorry) property.
So, for instance, the universal property (which you can translate as "the definition") of the coproduct $A\sqcup B$ (sorry, I don't like the notation $\oplus$ for the coproduct) of two objects $A$ and $B$ is the following: by definition it's an object $A\sqcup B$ together with two morphisms
$$
A \stackrel{\iota_A}{\longrightarrow} A\sqcup B \stackrel{\iota_B}{\longleftarrow} B
$$
satisfying the following property: whenever you have two morphisms
$$
f: A \longrightarrow C \qquad \text{and} \qquad g: B \longrightarrow C \ ,
$$
there is a canonical (unique) morphism
$$
h : A \sqcup B \longrightarrow C \qquad \text{such that} \qquad h\iota_A = f \quad \text{and} \quad h\iota_B = g \ .
$$
For instance, in the category of sets, the coproduct is the disjoint union, $\iota_A$ and $\iota_B$ are the inclusions and this morphism $h$ is
$$
h(x) =
\begin{cases}
f(a) & \text{if}\ x=a\in A \\
g(b) & \text{if}\ x=b\in B \ .
\end{cases}
$$
As for the product, its universal property (definition) is: the product of two objects $A$ and $B$ is an object $A \times B$ together with two morphisms
$$
A \stackrel{\pi_A}{\longleftarrow} A\times B \stackrel{\pi_B}{\longrightarrow} B
$$
satisfying the following property: whenever you have two morphisms
$$
f: C \longrightarrow A \qquad \text{and} \qquad g: C \longrightarrow B \ ,
$$
there is a canonical (unique) morphism
$$
h : C \longrightarrow A \times B \qquad \text{such that} \qquad \pi_A h = f \quad \text{and} \quad \pi_B h = g \ .
$$
For instance, in the category of sets, the product is the usual Cartesian product, $\pi_A$ and $\pi_B$ are the projections and this morphism $h$ is
$$
h(c) = (f(c), g(c)) \ .
$$
From these universal (sorry) properties follows, for instance, that whenever you have a morphism
$$
f: A \longrightarrow B
$$
you can "multiply" it by an object $X$, obtaining a morphism
$$
\mathrm{id}_X \times f : X \times A \longrightarrow X \times B \ .
$$
$\mathrm{id}_X \times f$ is the morphism induced by
$$
\mathrm{id}_X\pi_X : X\times A \longrightarrow X \qquad \text{and}\qquad f\pi_A : X\times A \longrightarrow B \ .
$$
and the universal (sorry) property of the product. For instance, in the category of sets
$$
(\mathrm{id}_X \times f) (x,a) = (x, f(a)) \ .
$$
So, your canonical morphism
$$
(X\times Y) \sqcup (X\times Z) \longrightarrow X \times (Y\sqcup Z)
$$
is deduced from the universal (sorry) properties defining product and coproduct and this last construction (also deduced from the aforementioned universal properties). Namely, to begin with, you have your morphisms that come together with the coproduct
$$
Y \stackrel{\iota_Y}{\longrightarrow} Y\sqcup Z \stackrel{\iota_Z}{\longleftarrow} Z \ .
$$
Next, you "multiply" them by $X$:
$$
X\times Y \stackrel{\mathrm{id}_X \times \iota_Y}{\longrightarrow}X\times ( Y\sqcup Z ) \stackrel{\mathrm{id}_X\times \iota_Z}{\longleftarrow} X\times Z \ .
$$
And now, the universal property of the coproduct gives you a canonical (unique) morphism
$$
\psi : (X\times Y) \sqcup (X\times Z) \longrightarrow X \times (Y\sqcup Z)
$$
such that
$$
\psi \iota_{X\times Y} = \mathrm{id}_X \times \iota_Y \qquad \text{and}\qquad \psi\iota_{X\times Z} = \mathrm{id}_X \times \iota_Z \ .
$$
In the category of sets, this $\psi$ is
$$
\psi (x, a ) =
\begin{cases}
(x, y) & \text{if}\ (x,a)=(x,y) \in X\times Y \\
(x, z) & \text{if}\ (x,a)=(x,z) \in X\times Z \ .
\end{cases}
$$
