Consider functions $g\in k[x,y]$ restricted to $\operatorname{Spec}(k[x,y]/(x^2,xy))$ with $k$ algebraically closed.(i.e. $Spec(k[x,y]/(x^2,xy))\subset \operatorname{Spec}(k[x,y])$ allows such restriction.) Denote $R=k[x,y]/(x^2,xy)$.
It is easy to check that stalk $R_m$ is nilpotent free if $m=(x,y-b),b\neq 0$.
The book claims that for $g|_m=g(m)+\frac{\partial g}{\partial y}(y-b)\in R_m$ for $m=(x,y-b)$ for $b\neq 0$.
$\textbf{Q:}$ Why $g|_m$ is determined up to $y-b$ order only here? My though process goes as the following. Since I am localizing away from $(x,y-b)$, certainly $y$ is invertible. So $R_m=(k[x,y]/(x^2,x))_{(x,y-b)}=(k[y])_{y-b}$. However, this says I can determine $g$ up to arbitrary order in $y$ direction.(i.e. I know $\frac{\partial^n g}{\partial y^n}|_{y=b}$ for any $n$.) Is this reasoning wrong here?
Ref. Mumford Algebraic Geometry 2, Chpt 2, Sec 3, Example 2.3.7