I'm currently trying to understand Silvermans example for the valuation on curves discussed in the answer to this post: Definition and example of "order of a function at a point of a curve"
Silverman has already shown that $M_P / M_P^2$ is generated by (y), but why does $M_P = (y)$ and then $ord_P(y) = 1$ follow from this fact?
This probably boils down to a deeper problem: Why are the elements if the function field with order one exactly the generators of $M_P$. I can see why $M_P = (t) \implies ord_P(t) = 1$ but I'm not really sure about the other implication. And how do the elements / generators of $M_P$ and $M_P^2$ relate to each other?
If $(R, \mathfrak{m}, k)$ is a noetherian local ring and $M$ is a finite $R$-module, a special case of Nakayama's lemma states that if $\mathfrak{m}M = M$ then $M = 0$.
In particular if $[m_1], \dots, [m_n]$ is a $k$-basis for $M/\mathfrak{m}M$, we can choose representatives $m_1, \dots, m_n \in M$. Consider the submodule $N \subset M$ generated by these representatives. Then, $\mathfrak{m}(M/N) = M/N$, since $M = N + \mathfrak{m}M$, so by Nakayama's lemma $M = N$ and $M$ is generated by $m_1, \dots, m_n$.
Now, in your situation, $M_p/M_p^2$ is generated by $[y]$, so $y$ lifts to a generator of $M_p$, and since $M_p \neq M_p^2$, we must have $y \notin M_p^2$. Hence $\operatorname{ord}_p(y) = 1$. Similarly if $\operatorname{ord}_p(t) = 1$ for some rational function $t$ then $t \in M_p \setminus M_p^2$, so $[t] \neq 0$ in $M_p/M_p^2$. As long as $\dim_k M_p/M_p^2 = 1$ (ie. the curve is nonsingular at $p$) it follows that $t$ lifts to a uniformizing parameter.