There are two aspects. First is that the notion of "ordered pair" is not the same as Kuratowski's version of the ordered pair, which is what you are asking about. Like many other things, the core notion is an abstraction (or interface), which specifies how the relevant objects must behave. For ordered pairs, we want them to have the following behaviour:
$∀x,y,z,w\ ( \ ⟨x,y⟩ = ⟨z,w⟩ ⇔ x = z ∧ y = w \ )$.
That is all we want. So mathematicians are perfectly fine with absolutely nothing more than this, treating ordered pairs as a primitive notion that satisfies the above abstract property. Ordered pairs are so crucial to mathematics that no mathematician would ever accept a foundational system for mathematics that cannot support this basic notion. Hence it is very important to know whether we can implement ordered pairs using whatever primitive notions we have available in any foundational system that we would like to use!
In some type theories, ordered pairs are primitive notions, so we do not have to look for an implementation. But in most presentations of ZFC set theory, the only primitive notion is the membership relation $∈$, and so we need to demonstrate a way of implementing ordered pairs. One way is Kuratowski's definition. That doesn't imply that ordered pairs are as Kuratowski's implementation; of course the abstraction is not the implementation. Besides, we could just as well define $⟨a,b⟩$ as $\{\{a,b\},\{b\}\}$, or as $\{\{\{\},\{a\}\},\{\{b\}\}\}$.
The second aspect is that whatever you were looking at failed to explain why $⟨x,y⟩ = ⟨z,w⟩$ implies $\{x\} = \{z\}$ and $\{y\} = \{w\}$. This is not as trivial as you might think. You need to be careful not to assume that $\{x,y\}$ has two (distinct) members, because it may be that $x = y$. To do this properly, you can first prove that given any $p = \{\{x\},\{x,y\}\}$ we have that $x$ is the unique member of both members of $p$, and then we have two cases:
- If $x = y$, then $p = \{\{y\}\}$, so for any $z$ such that $p = \{\{x\},\{x,z\}\}$ we have $\{y\} = \{x,z\}$ and hence $y = z$.
- If $x ≠ y$, then $\{x,y\}$ has two (distinct) members unlike $\{x\}$, so for any $z$ such that $p = \{\{x\},\{x,z\}\}$ we have $\{x,y\} = \{x,z\}$ and hence $y = z$.
I have left out a few tiny steps, but I hope you can carefully fill them in.