If $x=1$, then that means that $x^2 = 1$, also. Is the case the same if it has to do with lengths? That's if I got the $|x| = 1$ and then I raised to the power of two so $|x|^2= ...$ will it then be 2 or 1? I'm thinking as it has to do with length it might be addition instead of multiplication.
Where x is the length of a string.
A string is for example "abc" and it has the length 3, because it has 3 symbols in it. "a" has the length 1. So I'm wondering if |x| = |"a"|=1 then will |x|^2 be |"aa"|=2 or will it still be |"a"|=1.
$|x|$ is a number and squaring a number is a well defined and known operation. If $|x|=1$, then squaring the length of $x$ results again to 1, since $|x|^2 =1^2 = 1$.
Squaring the length of the string is not an operation on the string itself, but on its length.
The transformation that you describe in you example (from '$a$' to '$aa$') is a concatenation of two same strings '$a$' and '$a$'. In general, if $x$ and $y$ are strings with $|x|=n$ and $|y|=m$, then you can define a new string, $xy$ which is the concatenation of the two (writing first $x$ and then $y$ right after, without any separators, is the new string). The length of the new string, then, is $m + n$. For your example, consider $x = y =$ '$a$'.
In summary, for some string $x$ with $|x| = n$, we have $$|x|^2 = n^2$$ If by $x^2$ you mean $xx$, then also we have $$|x^2| = |xx| = 2|x| = 2n$$
Note that in the first case squaring is outside the bars ($|x|^2$), while in the second case the "squaring" notation (which actually denotes concatenation) is inside the bars ($|x^2|$).