Using the pumping lemma for a proof by contradiction
$u$ and $v$ are supposed to be strings such that $uv$ is a prefix of a
string in the language longer than $p$ symbols, $p$ being the pumping length. Furthermore, you do not
get to choose them. All you know is that their total length does not
exceed $p$ and they are prefix of a word $x$ in  the regular language, of length $n$.

You do not get to choose them because they are existentially
quantified in the statement of the pumping lemma ... you could choose
them if they were universally quantified. If you do not understand
this last remark, just ignore it.

What you can choose is the value of $n$ and the string $x=uvw$ of size $n$
that will be used for pumping.  You choose $n$ in relation to the
pumping length $p$ which depends only on the regular language.

I would suggest to choose $n$ quite big, so that you make sure $uv$ is
in the first half of the string $x$, since $| uv| ≤ p$.

I would also choose my string $x$ with an appropriate pattern of a and
b  ... and then check what happens when I pump. It may seem odd, but
you have to get even with that problem.