Perhaps one answer to your question about deformations is something like the following. A deformation over a complete local ring A (such as k[[t]]) is just a family X $\to$ Spec(A). Suppose that the fibers belong to some sort of moduli space M, such as the moduli space of curves. In the functorial point of view of moduli spaces, the family X $\to$ Spec(A) corresponds to a morphism Spec(A) $\to$ M that assigns to a point of Spec(A) the moduli of the fiber over this point. So, one parameter formal deformations (by this I just mean that A = k[[t]]) correspond precisely to the morphisms Spec(k[[t]]) $\to$ M. The scheme Hom(k[[t]], M) is called the space of arcs in M. If we fix the central fiber of the deformation then we get the space of arcs in M at the point corresponding to the central fiber. The space of arcs is a subtle and important invariant of a singularity. One can think of an arc (that is, a morphism Spec(k[[t]]) $\to$ M) as follows: if we had a curve in M then the arc would be the collection of jets this curve determines, ie all the derivatives of all orders of the curve (think of the way morphisms Spec$(k[t]/(t^2)) \to M$ determine the tangent vectors at the image of the closed point). So these deformations are telling us something significant about the local structure of the moduli space.

The construction of these one parameter formal deformations works regardless of the existence of any moduli space. It tells us what the space of arcs on the moduli space of whatever it is you are deforming should be.

algebraized. It would be easier to answer if you posted it as a separate question (and there are several people on MO who could give you good answers about it). Here I will just say that if C is any smooth curve over k, and you had a family over C, then looking at the formal n.h. of a point will give you something over k[[t]]. In other words, k[t] is not the unique way of algebraizing k[[t]]; any smooth curve will do. Thus you shouldn't prejudge the ... – Emerton Aug 1 '10 at 4:24