>To 3. What to do next: I believe that you need to multiply by some S to shift the valuation to what it needs to be to be a key polynomial and then fix the leading coefficient (because key polynomials are assumed to be monic.)

Here is my thought: Our goal is to find $\phi_{n+1}$ which $v_n$-divides $G$. By multiplying an $v_n$-unit with proper valuation on $G$, we may assume that $v_n(G)=0$. We factorize $G$ in $k_{v_n}$ and get a factor $\bar{\phi}_{n+1}$. By Theorem 13.1 of [Mac1936I], there exists a key polynomial $\phi_{n+1}$ satisfying that the reduction of $\phi_n^{\tau_n\flat_n\text{deg}\ \bar{\phi}_{n+1}}\phi_{n+1}$ to $k_{v_n}$ is $\bar{\phi}_{n+1}$. Here $\phi_n^{\tau_n\flat_n\text{deg}\ \bar{\phi}_{n+1}}$ is a $v_n$-unit (right?) to shift the valuation. Until now, everything is right?

As we have discussed, the first step of the construction in Theorem 13.1 of [Mac1936I] is lifting $\bar{\phi}_{n+1}$ to $\tilde{\phi}_{n+1}\in K[t]$. After that,
>multiply $\tilde{\phi}_{n+1}$ by $\phi_{n}^{\tau_n\sharp_n \text{deg}\ \bar{\phi}_{n+1}}$ and in the expansion of the resulting product drop all terms not of minimum value and then replace the leading coefficient of $\phi_n$ by $1$.

I'm not sure what's happening here. The term $\phi_{n}^{\tau_n\sharp_n \text{deg}\ \bar{\phi}_{n+1}}$ seems to be used to shift the valuation to $0$? The following steps are also confusing, especially
>replace the leading coefficient of $\phi_n$ by $1$.

Of course, it's used to make the polynomial monic. But why should I replace the leading coefficient of a key polynomial, instead of the expansion of the product?

P.S. Sorry for the poor math reading experience. I'm using a chrome extension (I modified it a little) to render MathJax. Next time I will try to use Github's math render, as you do.
P.P.S. Thanks for your patience. It's really helpful.