Intuition behind formal neighborhood and local ring and formal power series In The Geometry of Schemes by David Eisenbud and Joe Harris, on page 57, there is an explanation on "node" of a plane curve. The book says that, a curve $X\subseteq \mathbb A_{\mathbb C}^2$ has a node at $(0,0)$ if "the fibre product of $X$ with the formal neighborhood $\text{Spec }\mathbb C[[x,y]] \to \text{Spec }\mathbb C[x,y] =\mathbb A_{\mathbb C}^2$ is isomorphic to $\text{Spec }\mathbb C[[u,v]]/(uv)$".
I find it hard to picture this formal neighborhood ($[[]]$ denotes formal power series). It is explained in the book that the $\mathbb C[[]]$ neighborhood is smaller than $\text{Spec }\mathbb C[x,y],$ but how much smaller is it? I appreciate the formal neighborhood is "finer" than the local ring, because inverse of elements of $\mathbb C[x,y]$ outside the ideal $(x,y)$ can all be written as a formal power series. When passing to $\text{Spec},$ we reverse this inclusion.
On the other hand, however, I have some vague intuition which run into conflicts with what I write above. Firstly, $\text{Spec }\mathbb C[[x,y]]$ contains much more "curves" than $\mathbb A_{\mathbb C}^2,$ for example $y=e^x,$ and many of these curves gives prime ideal in $\text{Spec }\mathbb C[[x,y]].$ So apparently $\text{Spec }\mathbb C[[x,y]]$ is "larger", and the inclusion $\text{Spec }\mathbb C[[x,y]] \to \text{Spec }\mathbb C[x,y]$ is not injective, and extra strange curves get killed by this map. Secondly, why is $\text{Spec }\mathbb C[[x,y]]$ called a "neighborhood"? Apparently it contains some global structure (although I know "complex analytic functions" like this are restrictive), so it feels slightly unnatural to call it a neighborhood (after all, having the local picture of two curves intersecting at $(0,0)$ does not imply whether they intersect again somewhere else, but $\text{Spec }\mathbb C[[x,y]]$ seems to include how the curves behave on entire $\mathbb A_{\mathbb C}^2.$)
These conflicting thoughts means that, when the book states the phrase "isomorphic to $\text{Spec }\mathbb C[[u,v]]/(uv)$", I am still thinking about the possibility that we have another intersection point of the curve $X$ other than the origin, so how does the magic $\text{Spec }\mathbb C[[x,y]]$ isolates only one node of $X$ so that the other intersections disappear?
I know this might be a vague question, but to understand these things it just requires a little more explanation like this.
 A: First of all $\mathbb{C[[x,y]]}$ is local because it is only depend on $\mathbb{C}[x,y]_{(x,y)}$ because by definition it is $lim \frac{\mathbb{C}[x,y]}{(X,Y)^n}$ and all things outside $(X,Y)$ are invertible in this rings. more generally for the same reason completion of $A$ at the prime $P$ only depends on $A_P$. moreover as said in the comments if A is a quotient of $\mathbb{C}[X,Y]$ by some ideal $P$ containing $(X,Y)$ then completion of $A$ at $P$ is just $A\otimes \mathbb{C}[[X,Y]]$ so the creation is really saying that the type of singularity of $X$ at $x$ is given by $\hat{\mathcal{O}}_{X,x}$ which is obviously local.
as to why $\mathbb{C}[[X,Y]]$ is local one way to do this is to think about functions not points, if you have a smooth complex manifold every function locally at $0$ should be determined by its taylor expansion the fact that you also get taylor expansion of non algebraic analtyic function does not mean it is not local at zero, it means that you have an algebraic object that contains some extra local analytic date at $0$ and that's wonderful because you get an idea to do analysis when your base field is not $\mathbb{C}$.(and those extra points are not related to the global picture they are like $(x)$ generic points of the "subvarieties" passing through zero.)
another way is to look at points of this as a formal scheme: the ring $\mathbb{C}[[X,Y]]$ has a topology and it is a good idea to look only at the prime ideals that are open with respect to that topology, in this way you get rid of those "weird" points.
there is also a more algebraic explanation $\mathbb{C}[[X,Y]]$ is a neighborhood of $0$ in the faithfully flat topology but I thing you prefer the geometric arguments.
