I have recently learned about uniform spaces. My intuition is that uniformity is about notion of closeness (points can be quite close, or very close, or very very close.. to each other). However, I fail to grasp some intuition of this notion.
My questions:
Why is uniform space defined with the product of spaces? What do the axioms say informally?
Why is completeness a uniform-invariant property? (Not changed under uniform isomorphism)
Disclaimer: Uniformity = uniform space = space with uniform structure.
When looking for intuition on more general concepts it is often useful to take a step back and apply them to the structures that they generalize. In this particular case it is useful to look at metric spaces.
Since the first question was already answered in the comments I will focus on the first.
Let $X$ be a metric space. Then the filter basis that generates the uniformity (this is a filter) that coincides with the topology of $X$ is generated by the entourages
$U_{\epsilon} = \left \{ \left ( x,y \right ) : d\left ( x,y \right )\leq \epsilon \right \}$.
Focusing on the filter basis elements we see that for a fixed $\epsilon$ the pairs contained in $U_{\epsilon}$ are precisely those within $\epsilon$ of each other. It is clear that we can also think of $U_{\epsilon}$ as a union of the sets (union over all of $x \in X$)
$U_{\epsilon}^{x} = \left \{ y : d\left ( x,y \right )\leq \epsilon \right \}$.
I hope that gives you some basic intuition about what entourages can be.
The reason why everything is formulated on the product $X \times X$ is because uniformities encode relationships between points of a topologicial space. Topological spaces without a uniform structure do not formalize the notion of closeness of points. Recall that the metric is also a function from the product space $X \times X$ to $\mathbb{R_{0}^{+}}$. So working on $X \times X$ is necessary.
In fact this can be explained even more clearly using some of the axioms for the uniformity. Let $U$ be a uniformity. Then, by definition there exists another uniformity $V$ such that $V \circ V \subset U$. We can show that in a metric space this follows from the triangle inequality (it can be seen as a generalization). Since the elements $U_{\epsilon}$ are a filter basis then there exists a $\epsilon>0$ such that
$U_{\epsilon} \subset U$.
Then the triangle inequality (check this) implies that
$U_{\epsilon/2} \circ U_{\epsilon/2}\subset U_{\epsilon} \subset U$.
You can continue and draw other useful parallels between the remaining uniform properties and their corresponding metric ones like the connection of the symmetry property of a metric and
$U$ is an entourage implies $U^{-1}$ is an entourage.
I hope you will find the insight helpful.