We have the following inequality
here $S_n$ is a simple symmetric random walk, and $b_n=[2n\log\log n]^{1/2}$, and $\xi(m)\rightarrow 1$. I don't know how we get the two orange parts. For the first one I really have no idea, for the second one I tried \begin{align} \frac{\alpha b_{\alpha^m}}{\sqrt{\alpha^{m+1}}}&=\frac{\alpha}{\sqrt{\alpha^{m+1}}}\left[2\alpha^m \log\log \alpha^m\right]^{1/2}\\ &=\alpha\left[\frac{2\log\log \alpha^m}{\alpha}\right]^{1/2}\\ &=\alpha\left[\frac{2\log (m\log \alpha)}{\alpha}\right]^{1/2} \end{align} which doesn't give us the $\xi(m)$.
First one is a lemma in the notes (Lemma 5.1), which, by using reflection principle, links the probability of a $\sup$ of a random walk up to time $n$ by the probability that it will end up in a particular place at time $n$.
For the second part, you are almost there: just write $\xi(m) = \frac{\log m + \log \log \alpha}{\log m}$.