Great question! I don't actually like this characterization of the product of two spaces all that much; to me, it seems to be hiding the real reason we are interested in this space, which is that it is the category-theoretic product of these spaces.
What does this mean? Well, first of all, ask yourself a question about sets, rather than topological spaces. If $A,B,C$ are sets, what does it mean to give a function from $C\to A\times B$?
After a bit of thought, you realize that specifying a function from $C\to A\times B$ is exactly the same as specifying a pair of functions $f\colon C\to A$ and $g\colon C\to B$. We may write the function as $(f,g)\colon C\to A\times B$, sending $c$ to $(f(c),g(c))$.
If we like, we can recover the functions $f$ and $g$ from the funciton $(f,g)$ by composing it with the projection maps $\Pi_1\colon A\times B\to A$ and $\Pi_2\colon A\times B\to B$. Put another way, if $h\colon C\to A\times B$ is any function, then we always have $h=(\Pi_1\circ h, \Pi_2\circ h)$.
The idea for the product of too topological spaces is the same, except now we replace 'set' with 'topological space' and 'function' with 'continuous function'.
Proposition If $X,Y,Z$ are topological spaces and $h\colon Z\to X\times Y$ is continuous, then $\Pi_1\colon X\times Y\to X$ and $\Pi_2\colon X\times Y\to Y$ are continuous and $h=(\Pi_1\circ h, \Pi_2\circ h)$.
In other words, continuous functions into a product space correspond precisely to pairs of continuous functions going into each of the two spaces.
What has this to do with your proposition? Well, in category theory, we have something called a limit, and a product is a particular special case of a limit. A beautiful fact about topological spaces is that we may always construct limits by taking the appropriate limit in the category of sets, and then endowing this set with the coarsest possible topology such that the appropriate functions are continuous.
If you don't know category theory, that paragraph probably didn't make much sense, so I'll try to illustrate it in this case.
We start with two topological spaces $X$ and $Y$ and we want to construct the product space so that the proposition above is satisfied. First, we pass to the underlying sets $X$ and $Y$ of these topological spaces. We already know how to form the product of these sets - it is just the set $X\times Y$ of pair $(x,y)$. Now we want to know what topology to put on this set.
Firstly, we are going to need $\Pi_1$ and $\Pi_2$ to be continuous. This means that $\Pi_1\circ h$ and $\Pi_2\circ h$ are automatically going to be continuous. In order to make the proposition hold, we need the maps $(f,g)$ to be continuous whenever $f,g$ are continuous.
The coarser the topology we put on $X\times Y$, the 'easier' it will be for a function into $X\times Y$ to be continuous, since there will be fewer open sets to worry about. And it turns out that if we take the coarsest topology of all (subject to $\Pi_1,\Pi_2$ being continuous) then the proposition holds.