Let $X,Y$ be curves of $\mathbb{A}^2$ given by irreducible polynomials $f,g$ respectively, where the ground field $k$ is algebraically closed. Then it is known that the dimension of $k[x,y]/(f,g)$ as a $k$-vector space is equal to the cardinality of $X \cap Y$.
Now let $P \in X \cap Y$. The intersection multiplicity of $X,Y$ at $P$ is defined as the dimension of the $k$-vector space $\mathcal{O}_P/(f,g)$, where $\mathcal{O}_P$ is the local ring at $P$. Intuitively, the effect of localization is that of removing contributions of points in $X \cap Y$ other than $P$. Question 1: How can we see that rigorously?
In particular, it is mentioned here (top of page 3) http://www.math.lsa.umich.edu/~hochster/cmrvw.ps that "it suffices to invert enough elements of $k[x,y]$ such that for any point in $V(f,g)-P$, at least one of the inverted elements vanishes at that point. Question 2: Why is it enough to do that? I.e. why inverting such elements removes the effect of other points in $X \cap Y$?
Question 3: How can we see that $\dim_k \mathcal{O}_P/(f,g) \ge \mu_P(X) \mu_P(Y)$? This is problem I.5.4.(a) in Hartshorne and i have found none of the available online solutions satisfactory.
Edit:
Definition: $\mu_P(X)$ is the multiplicity of point $P$ in the variety $X$. It can be computed as follows: do a change of coordinates such that $P$ maps to $(0,0)$ and write $f=f_r+f_{r+1}+\cdots+f_n$, where $f_i$ is homogeneous of degree $i$. Then $\mu_P(X)=r$.