On local rings of a normalization Let $X$ an irreducible singular curve with a singular point $x$. Consider $A$, the normalization of $\mathcal{O}_{X,x}$, and $x_1,\dots,x_r$ the points over $x$ in the normalization of $X$. Why $A\cong \mathcal{O}_{x_1}\cap\dots\cap\mathcal{O}_{x_r}$?
Moreover assume that the curve is over a field $k$. Why do the dimension of $A/\mathcal{O}_{X,x}$ is $r-1$?
 A: The comment by Matt might work here, but there is a more general fact at play, that is also valid for higher-dimensional varieties:
Theorem [1, Thm 4.7] Let $A$ be an integral domain with field of fractions $K$. We consider any localization of $A$ as a subring of $K$. Then in this sense we have $$A = \bigcap_{\mathfrak{m} \subset A} A_\mathfrak{m},$$ where the intersection is over all maximal ideals $\mathfrak{m} \in A$
Now in your example, the maximal ideals $\mathfrak{m} \subset A$ correspond exactly to the points $x_1, \dotsc, x_r$, and localizing $A$ with respect to them gives $\mathcal{O}_{x_i}$.
Your second claim that $\dim A / \mathcal{O}_{X,x} = r - 1$ is not true. For example consider the cusp
$$ X = \operatorname{Spec}(k[x,y] / (x^2 - y^3).$$
The normalization of $X$ is just $\tilde X = \mathbb{A}^1$, and the map $\tilde X \to X$ is given by
\begin{align}
\mathbb{A}^1 & \to X \subset \mathbb{A}^2 \\
t & \mapsto (t^3, t^2).
\end{align}
On coordinate rings the map is given as $x \mapsto t^3, y \mapsto t^2$. The only value for $t$ that hits the singular point $x = (0,0)$ is $t = 0$, so $r=1$ here. But clearly $A / \mathcal{O}_{X,x} \neq 0$, because $x$ is a singular point of $X$.
[1] Matsumura, Commutative Ring Theory
