The categorical definitions of the basic objects (products, coproducts, equalizers, coequalizers, limits, colimits, etc) arise from abstracting a particular situation that shows up in many instances.
The notion of a "product" shows up repeatedly among Sets (cartesian product), Groups (cartesian product again), Rings (cartesian product yet again), Modules (and again), Topological spaces (and again).
Their "object" properties vary from category to category; in the class of abelian groups, the direct product contains the direct sum, which is an important objects in its own right (it's also important in Groups, but less so). Among Unital Rings, however, the direct sum is not an interesting object when it is different from the product (because it is not a unital ring). In topological spaces, there is a "natural" way of defining the topology on a product (the "box" topology), but it turns out to be lacking in some uneasy sense when there are too many factors in the product... and there is an alternative possibility (the "product topology") that seems to be behave better...
The general, guiding principle in Category Theory, though, is that what is important is not what an object "is", but rather what it "does": how it behaves relative to other objects in terms of arrows/morphisms. So we want to abstract the properties that all those instances share in $\mathbf{Sets}$, $\mathbf{Groups}$, $\mathbf{AbGroups}$, $\mathbf{Rings}$, $\mathbf{UnitalRings}$, $\mathbf{Semigroups}$, $\mathbf{TopSpaces}$, etc.
(In the latter case, it even suggests which of the two possible topologies "should" be used).
So... what do all these specific instances all have in common? From the cartesian product we can map to each factor: if we view the cartesian product $\times X_i$ as the set of functions $f\colon I\to \cup X_i$ with $f(i)\in X_i$ for each $i\in I$, then we have a natural map from the product to each $X_i$ by "evaluation": $\pi_i\colon f\longmapsto f(i)$. And $f$ is completely determined by these images; so that if you "select" $x_i\in X_i$ for each $i$, then this gives you a unique $\mathbf{x}\in \times X_i$.
Ah, but again: we don't want to think in terms of elements (what the objects "are"), but rather in terms of maps. For $\mathbf{Sets}$, selecting objects is the same as mapping from single element sets... but again we are worrying about the what the objects "are" (single element sets)...
Okay: since we care about maps, note this: not only can we go from $\times X_i$ to each $X_i$. These maps are such that if you have any set $Y$, and any maps $g_i\colon Y\to X_i$, then each $y\in Y$ determines an element of $\times X_i$, namely, $y\colon i\to g_i(y)$. Put another way:
If we let $\pi_i\colon\times X_i\to X_i$ be the "evaluate at $i$" map, then for any set $Y$ and any maps $g_i\colon Y\to X_i$, there is a way to map $g\colon Y \to \times X_i$ in such a way that $\pi_i\circ g(y) = g_i(y)$ for every $i$.
What's more, $g$ is completely forced by this condition. This is a good "categorical" property, because it is given entirely in terms of morphisms.
Does this property "go through" the other instances? The group with the coordinatewise product? The semigroup/ring/module with coordinatewise product? Yes! Good. What about topological spaces? Yes with respect to the product topology... no with respect to the Box topology... so we want to use the "product topology" for the "product".
This definition matches the many instances, and has nice "categorical" properties. It completely characterizes the product up to unique isomorphism. So it seems like a good definition. Turns out to be the right way to generalize the properties of the cartesian product/direct product to other categories.
Writing $(\Delta \mid F)$ is a slight abuse of notation. Remember that $\Delta$ is a functor $\mathcal C \to \mathcal C^{\mathcal C'}$, while $F$ is a functor $\mathcal C' \to \mathcal C$.
To form a comma category $(G \mid H)$, the functors $G$ and $H$ need to have the same codomain.
So, we actually consider $F$ as a functor $\bf1 \to \mathcal C ^{\mathcal C'}$. Let's write it as $\hat F :\bf 1 \to \mathcal C ^{\mathcal C'}$, where $\hat F \star = F$ (and $\star$ is the unique object of $\bf 1$).
Now we see that the objects of $(\Delta \mid \hat F)$ are triples $(A, \star, h)$, where $A$ is an object of $\mathcal C$ and $h$ is a morphism $\Delta A \to \hat F\star = F$. So $h$ is a natural transformation between the functors $\Delta A$ and $F$.
It is not hard to verify that this is the same thing as a cone from $A$ to $F$.
Edit: By “cone to $F$”, I mean an object $A$ of $\mathcal C$, and a family of morphisms $\gamma_X :A \to F X$, indexed by objects $X$ of $\mathcal C’$, such that for every $f : X \to Y$, we have $\gamma_Y = Ff \circ \gamma_X$.
A morphism of cones $(A, \{\gamma_X\}_X) \to (B, \{\delta_X\}_X)$ is a morphism $f : A \to B$ such that for every X we have $\gamma_X = \delta_Y \circ f$.
So given a category $\mathcal C$ and a functor $F : \mathcal C’ \to \mathcal C$ we can form the category of cones to $F$ in $\mathcal C$. This is isomorphic to the comma category $(\Delta \mid \hat F)$.
Best Answer
Here is the short answer: The conditions which you describe make sure that $\mathcal C(z,a\times b) \to \mathcal C(z,a) \times \mathcal C(z,b)$, $f \mapsto (\pi_af,\pi_bf)$ is an isomorphism for all $z$. This is why you can say sentences like: A map into $a\times b$ is the same as a pair of maps (one into $a$ and one into $b$).
Here is the long version. I assume that you know about the Yoneda lemma. A lot of interesting definitions in category theory are actually representations of $\mathtt{Set}$-valued functors. A representation of a contravariant functor $F: \mathcal C \to \mathtt{Set}$ is an object $c$ in $\mathcal C$ together with a natural isomorphism $\alpha: \mathcal C(-,c) \to F$ of functors. A representation usually tells us that the maps into $c$ have a nice description, namely $F$.
For example consider $\mathcal C = \mathtt{Set}$ and $c = \{0,1\} = \Omega$. Then the set maps from $A$ into $\Omega$ correspond naturally to the subsets of $A$, and $\mathtt{Set}(-,\Omega)$ is naturally isomorphic to the power set functor.
There are many (!) examples of representable covariant and contravariant functors in category theory. You could define the product of two objects $a$ and $b$ as follows. It should be an object $a\times b$ such that maps into $a\times b$ correspond naturally to pairs of maps into $a$ and $b$. In other words you like to have a representation $\alpha: \mathcal C(-,a\times b) \to \mathcal C(-,a) \times \mathcal C(-,b)$. This determines $a\times b$ up to unique compatible isomorphisms.
Now here is the catch. Because of Yoneda's lemma giving a natural transformation $\alpha: \mathcal C(-,c) \to F$ is the same as specifying an element $\alpha(1_c)\in Fc$, and $\alpha$ is an isomorphism if and only if $\alpha(1_c) \in Fc$ is terminal in the category of elements of $F$. Now in the case of a representation $\mathcal C(-,a\times b) = \mathcal C(-,a) \times \mathcal C(-,b)$ this means that specifying the natural isomorphism is the same as giving a pair of morphisms $(\pi_a,\pi_b) \in \mathcal C(a\times b,a)\times \mathcal C(a\times b,b)$. The condition that the transformation is an isomorphism becomes the universal property of the product, which you describe in your question.