Local ring of a projective curve at a point and order of a rational function at that point This is from Silverman's Arithmetic of Elliptic Curves. I'm confused by some of the definitions. Let me first lay down the relevant definitions.
Let $V$ be an affine variety. The ideal of $V$ is $I(V) = \{f\in \bar K[X] \mid f(P)=0\;\forall P\in V \}$. If $I(V)$ is generated by elements in $K[X]$, the $V$ is said to be defined over $K$ and define $I(V/K) = I(V)\cap K[X]$. Define the coordinate rings 
$$
\bar K[V] = \bar K[X] / I(V) \quad \text{ and } \quad K[V] = K[X]/I(V/K).
$$
The function field $K(V)$ is the field of fractions of $K[V]$, and the function field $\bar K(V)$ is the field of fractions of $\bar K[V]$.
Let $P$ be a point on the affine variety $V$. The ideal $M_P$ in the coordinate ring $\bar K[V]$ of $V$ is defined by $M_P = \{ f\in \bar K[V]\mid f(P) = 0\}$. So $M_P$ is a maximal ideal in $\bar K[V]$ since $\bar K[V]/M_P$ is isomorphic to the field $\bar K$ via $f \mapsto f(P)$.
The local ring of $V$ at $P$, denoted by $\bar K[V]_P$, is 
$\bar K[V]_P = \{ F \in \bar K[V] \mid F =f/g$ for some $f,g \in \bar K[V]$  with $g(P) \neq 0\}$.
Change of notation. Let $V$ be a projective variety and $P$ be a point on $V$. The function field $K(V)$ is defined to be $K(V\cap \mathbb{A}^n)$. Similarly, $\bar K(V) = \bar K(V\cap \mathbb{A}^n)$. Here the author notes that it doesn't matter which $\mathbb{A}^n$ we choose: for different choices of $\mathbb{A}^n$, the different $K(V)$ are canonically isomorphic. I'm guessing this canonical isomorphism is defined by homogenizing an element of one $K(V)$ and then dehomogenizing it to another $K(V)$???
Choose an affine space $\mathbb{A}^n$ such that $P \in \mathbb{A}^n$. The local ring of $V$ at $P$ is the local ring of $V\cap \mathbb{A}^n$ at $P$.
Let $C$ be a curve, i.e. projective variety of dimension 1.
It's noted above that $M_P = \{ f\in \bar K[C]\mid f(P) = 0\}$ is a maximal ideal in $\bar K[V]$. But the same $M_P$ is also the unique maximal ideal in $\bar K[C]_P$. (Am I right here?)
Now the confusing part. 
Let $P$ be a smooth point on a curve $C$. The (normalized) valuation on $\bar K[C]_P$ is 
$$
\mathrm{ord}_P: \bar K[C]_P \to \{0, 1, 2,\dots \} \cup \{\infty\},\\
\mathrm{ord}_P(f) = \mathrm{sup}\{d\in\mathbb{Z}\mid f \in M_P^d \}.
$$
Using $\mathrm{ord}_P(f/g) = \mathrm{ord}_P(f) -\mathrm{ord}_P(g)$, we extend $\mathrm{ord}_P$ to $\bar K(C)$, $\mathrm{ord}_P: \bar K(C) \to \mathbb{Z}\cup\{\infty\}$.
My questions: $M_P^d$ is a subset of $\bar K[C]$, but $\mathrm{ord}_P$ (before extending) is defined on $\bar K[C]_P$, which is a subset of $\bar K(C)$. So how does the initial defn of $\mathrm{ord}_P$ work?
What does it mean for $\mathrm{ord}_P(f) = 0$ or $\mathrm{ord}_P(f) = \infty$? I don't know what $M_P^0$ or $M_P^\infty$ even means.
The next defn is more confusing. 
Let $f \in \bar K(C)$. If $\mathrm{ord}_P(f) > 0$, $f$ has a zero at $P$. If $\mathrm{ord}_P(f) < 0$, $f$ has a pole at $P$. If $\mathrm{ord}_P(f) \ge 0$, $f$ is regular (defined) at $P$ and we can evaluate $f(P)$. Otherwise, $f$ has a pole at $P$ and we write $f(P) = \infty$.
The part about $f$ having a zero or is regular at $P$ seems to me like a theorem/proposition rather than a defn. In fact, in proposition 2.1, the author seems to use the fact that if $\mathrm{ord}_P(f) = 0$, where $f\in \bar K(C)$, then $f(P)$ is defined and $f(P) \neq 0$.
Here a rational function $f \in \bar K(C)$ is defined/regular at $P$, by defn, if $f \in \bar K[V]_P$, i.e. $f$ has a representative whose denominator is non-zero at $P$.
 A: Firstly, on the canonical isomorphism for different choices of $\mathbb{A}^n$: observe that for two different choices of $\mathbb{A}^n$ you find a point $P$ in the intersection of those two choices which is contained in an affine variety openly contained in the intersection of those two choices. Thus, for each choice the local rings at $P$ are canonically isomorphic. Then convince yourself that $K(V)$ is just the fraction field of $K[V]_P$ for any point $P \in V$. 
Second: the ring $K[V]_P$ is the localization of $K[V]$ at the maximal ideal $M_P \subseteq K[V]$. Thus, the maximal ideal in $K[V]_P$ (which we also denote by $M_P$) is given as $\{ \frac{f}{g} \mid f \in M_P \subseteq K[V], g \in K[V] \backslash M_P \} = \{\frac{f}{g} \mid f(P)=0, g(P) \neq 0\} $. Think of it as the functions on $V$ that are defined at $P$ and vanish at $P$.
Thirdly: in the definition of the order function, by $M_P$ we mean the unique maximal ideal in $K[C]_P$ (which is just the localization of $M_P \subseteq K[C]$ in $K[C]_P$).
Fourthly: we have for a ring $R$ and $I \subseteq R$ an ideal, one sets $I^0 = R$. Thus, $M_P^0 = k[C]_P$. That means, $f \in K[C]_P$ is of order $\mathrm{ord}_P(f)=0$ if and only if $f \in K[C]_P$ is a unit (i.e. not contained in the maximal ideal of the local ring), or, in other words, if $f$ does not vanish at $P$ (n.b., by definition then $\frac{1}{f}$ is also an element of $K[C]_P$).
Fiftly, $\mathrm{ord}(f)= \infty$ just means $f \in M_P^k$ for all $k \geq 0$. But in a discrete valuation ring we have $\bigcap_{k \in \mathbb{N}} M^k=0$. So $\mathrm{ord}(f) = \infty$ is true if and only if $f=0$.
Sixthly, the definition on poles and zeroes actually only is a definition. A very geometrically intuitive one though (and that is why it might seem to you that there is some statement hidden in there).
Seventhly, let us check that $\mathrm{ord}_P(f) = 0$ if and only if $f(P)$ is defined and $f(P) \neq 0$ for $f \in K(C)$. Pick a uniformizer $t$ at $P$ (i.e. an element $t$ generating the maximal ideal $M_P$). Write $f= g/h$ with $g,h \in K[C] \subseteq K[C]_P$. Then $\mathrm{ord}_P(g) = \mathrm{ord}_P(h)$, so we have $g= g't^m$, $h=h't^m$ for some $m=\mathrm{ord}_P(g)$ and $g', h'$ of $\mathrm{ord}_P(g')=\mathrm{ord}_P(h')=0$. But then, $f= g'/h'$, and by (Fourthly) we see that $1/h' \in K[C]_P$, so we get that $f \in K[C]_P$. On the other hand, $g' \notin M_P$ (because $\mathrm{ord}_P(g')=0$), so $f(P) \neq 0$.
