I'm reading this wikipedia article about infinitely near points.
In the section "Applications", the article says: If $C,D$ are irreducible curves on a smooth surface $S$ which intersect in a point $p$, then: $$m_p(C\cap D)=\sum_{x\text{ infinitely near }p}m_x(C)m_x(D)$$
Before looking for a proof of this formula, I tried to see it working in a concrete example.
I took the curves $C: y^2=x^2(x+1)$ and $D: y=x$ in $\Bbb{A}_{(x,y)}^2$ and looked at the blow up $X\subset \Bbb{A}^2_{(x,y)}\times\Bbb{P}^1_{(s:t)}$ defined by $sy=tx$.
In this case, $m_p(C\cap D)=\dim_k\mathcal{O}_p/(y^2-x^2(x+1),y-x)=\dim_k\mathcal{O}_p/(y^3,x)=3$.
Now locally in $\{s\neq 0\}$ we write $u:=\frac{t}{s}$, so that the strict transforms $\widehat{C},\widehat{D}$ look like $u^2=x+1$ and $u=1$ respectively. In particular $\widehat{C},\widehat{D}$ meet the exeptional divisor at $u=\pm 1$ and $u=1$ respectively. The computation in $\{t\neq 0\}$ shows nothing new.
This means the only infinitely near point of both $C,D$ is $x:=((0,0),(1:1))$, which is a simple point in both $\widehat{C},\widehat{D}$. Therefore: $$\sum_{x\text{ infinitely near }p}m_x(C)m_x(D)=1\cdot 1=1\neq 3$$
What am I missing here?
You have forgotten to consider the infinitely near point of order zero: $p$ itself. In which case, $m_p(C) = 2$ and $m_p(D) = 1$; hence,
$$ m_p(C)m_p(D) + m_x(\hat{C}) m_x(\hat{D}) = 2\cdot 1 + 1\cdot 1 = 3 = m_p(C\cap D).$$
(It's likely worthwhile to also mention that there are no further higher order infinitely near points since the tangent lines of the strict transforms at $x$ are distinct; and hence, the strict transforms of $\hat{C}$ and $\hat{D}$, in the blowup of $x$, do not meet in the exceptional curve.)
As an aside, it seems best to regard this formula as an explanation of the difference between the multiplicity of the intersection and the product of the multiplicities: $m_p(C\cap D) - m_p(C)m_p(D)$. Under this view, the importance of considering the point $p$ in the product of the multiplicities is more readily apparent.