let me ask you a question which will show my poor understanding of stalks and ringed spaces.. I hope that this example will help me clarifying the subject. So here we go: I've read (in particular from Michel Brion's "Local properties of algebraic group actions", example 1.12) that "a cuspidal curve $X$ can be obtained from $\mathbb{P}^1$ by sending the fat point $Spec(\mathcal{O}_{\mathbb{P}^1,\infty}/\mathfrak{M}^2$) to the cusp $x$".
I have a hard time understanding the meaning of this sentence. In particular:
1) Does it mean that $\mathcal{O}_{\mathbb{P}^1,\infty}/\mathfrak{M}^2$ is isomorphic to $\mathcal{O}_{X,x}$, with $X$ the cuspidal curve? I can't really see it, because, given $ZX^2=Y^3$ the homogeneous equation of $X$ in $\mathbb{P}^2$, with cusp in $x=[0,0,1]$, then $\mathcal{O}_{X,x}=\{\frac{F}{G}|F,G\in \mathbb{C}[X,Y,Z]/(ZX^2=Y^3),G([0,0,1])\neq 0\}$
2) Is it true that, away from the cusp $x$, the stalks $\mathcal{O}_{\mathbb{P}^1,t}$ and $\mathcal{O}_{X,y}$ are isomorphic?
No, I think a different construction is meant here. Let me start with a case, that is easier to picture: Given an abstract curve $C$ and two points $P_1$ and $P_2$ on it, we can form a new (singular) curve $C'$, where we take as a topological space the one obtained by gluing the points and as a structure sheaf $\mathcal{O}_{C'}(U) = \{f \in \mathcal{O}_C| f(P_1) = f(P_2)\}$. Now your case is similar, but instead of gluing two separate points, we contract a double point on $\mathbb{P}^1$. That is, we define a new curve, where we e.g. contract the fat point $\mathbb{C}[t]/(t^2)$ to a closed point and take as new structure sheaf $\mathcal{O}_{C'}(U) = \{f \in \mathcal{O}_C| f'(0) = 0\}$.
To convince yourself, that we get the cuspidal curve $C$ by this procedure, consider the birational map (in affine coordinates): $\mathbb{A}^1 \rightarrow C$, $t \rightarrow (t^3, t^2)$. It is bijective and gives as always a map on coordinate rings: $\mathbb{C}[x,y]/(x^2-y^3) \rightarrow \mathbb{C}[t]$, $ p(x,y) \rightarrow p(t^3, t^2)$. On stalks away from zero this will be an isomorphism, but over zero it is not surjective, since the polynomial $t$ is not in its image. Removing $(t)$ as we did above fixes this (note that the requirement above $f'(0) = 0$ exactly says, that $f$ as a sum of monomials has no linear component).