Why does the Big O notation of summation give log I’m not understanding the following line in a proof
I believe it should be $O(h^2\psi(2h))$ since $\sum_{j=1}^{h}\frac{1}{j}\leqslant h\cdot1=h$

You found the correct statement
$$\sum_{j=1}^h \frac1j \in O(h).$$
The authors of the paper used the correct statement
$$\sum_{j=1}^h \frac1j \in O(\log h),$$
as explained in the answer by Parcly Taxel. Since $\log h \in o(h)$, the paper uses the sharper, "better" bound. Yours isn't wrong, it's just not as good as could be.
You can find details about why $\sum_{j=1}^h \frac1j \in O(\log h)$ in the wikipedia entry for the Harmonic numbers.
