Let $f\colon X \to Y$ be a map of schemes. For this to be a closed immersion, the sheaf-level requirement is that $f^\sharp\colon \mathcal{O}_Y \to f_\ast\mathcal{O}_X$ needs to be a surjective map of sheaves.
In practice we often want to check surjectivity on stalks. Now there are two possible maps on stalks, and I'm really confused whether surjectivity for the two maps is equivalent, and couldn't find any definitive source anywhere.
On the one hand, there is the usual induced map on stalks $f^\sharp_q\colon \mathcal{O}_{Y,q} \to (f_\ast\mathcal{O}_X)_q$ for all $q\in Y$. There is also the map $\phi_p\colon \mathcal{O}_{Y, f(p)} \to \mathcal{O}_{X, p}$ for all $p\in X$.
Now I understand that $f^\sharp_q$ being surjective is equivalent to $f^\sharp_q$ for all $q\in Y$. But in some proofs I've seen, people seem to assume that $\phi_p$ being surjective for all $p\in X$ is also an equivalent condition.
Question. (i) Is this second equivalence true? Does $\phi_p$ surjective implies $f^\sharp_q$ surjective, or the other way around? (ii) Is $(f_\ast\mathcal{O}_X)_q = \mathcal{O}_{X, p}$ when $f(p)=q$?
Contexts & References (feel free to ignore).
There is a related question here which refers to Qing Liu's Algebraic Geometry and Arithmetic Curves, proof of Prop 2.24, which claims (ii) of my question to be true. But I can't see why it holds, when $q\in f(X)$: I don't think the argument for this case given in that question is correct.