If a one-way functions (OWF) exist, then there exits a OWF that is computable in quadratic running time by a padding argument
When we talk about quadratic running time, we mean $O(n^2)$ running time where $n$ is the length of the input -- i.e., the running time is a constant times the square of the length of the input.  Make sure you understand what the length of the input is here.  By padding the input, we've made the input longer.  Even though the overall running time doesn't change, the size of the input is larger, so relatively speaking, the running time is now a slower-growing function of the input size.

For instance, suppose $f$ can be computed in cubic time: in $O(k^3)$ time, where $n$ is the length of the input to $f$.  Now consider the padded function $f'$, which has been padded out to the length $k^{1.5}$.  What is the running time of $f'$?  Its running time remains $O(k^3)$, but now the input is of size $k^{1.5}$.  Re-expressing the running time of $f'$ as a function of its input size, we let $n=k^{1.5}$ be the length of the input to $f'$, and then notice that $k^3=n^2$, so the running time of $f'$ is $O(n^2)$: quadratic in the size of its input.



On your specific question: yes, the $O(n^2)$ seems unnecessarily loose.  Your argument that the running time of $f'$ is $O(n)$ (linear in the length of its input) looks right to me.  I don't know why they mentioned $O(n^2)$ as opposed to $O(n)$.