I am following the two examples in Masulli's document, and even though everything is very clear I can't figure out a small detail.
In page 25 he goes: It can be shown that the vertical lines in the diagram represent multiplication by $p = 2$ in $_2π_*^S$. Later on, he also says that they're also multiplication by $h_0$.
What I don't see is why multiplication by $h_0$ is equivalent to multiplication by $2$. Is it always true (when in the Adams spectral sequence we take $X = \mathbb{S}$ the sphere spectrum)? If it's nontrivial, is there some place where it's proven?
In the case of the spectrum $KO$ it turns out to be also satisfied (see page 32).
Yes, it's true that $h_0$ realizes to $2$ in general. It suffices to prove this for $X = S$. The easy argument would be to simply note that $\pi_0^S \cong \mathbb{Z}$, so on the $0$-stem of the $E_\infty$ page we have a filtration of $\mathbb{Z}$ where all the filtration quotients are $\mathbb{Z}/2$. This filtration must be $$\mathbb{Z} \supset 2\mathbb{Z} \supset 4\mathbb{Z} \supset \cdots.$$ Under this identification, we see that $h_0 \in E_\infty^{1,1}$ corresponds to $2 \in \pi_0^S$.