I presume that by $\widehat{\mathcal{O}}_{X, V}$ you mean the ring obtained by completing the local ring at the generic point of $V$ at its maximal ideal.

You're right that your description ($\widehat{\mathcal{O}}_{X, V}=K(V)[[x_1, ..., x_k]]$) follows immediately from the Cohen structure theorem (see e.g. 10.149.8 here); in fact it is not necessary that everything be smooth, just that the local ring at the generic point of $V$ is regular. Indeed, any complete regular local ring containing a field is a power series ring over a field. In this case, your ring contains $K(V)$, and $K(V)$ is the maximal field it contains (because it is isomorphic to its residue field, for example). This is prove on page 11 here, for example. The fact that $k$ equals the codimension follows by considering the Krull dimension of the local ring at the generic point of $V$.

But I suspect that this is not what you mean to look for; this is something like "an infinitesimal neighborhood of the generic point of $V$". I suspect you really mean to look at the formal scheme obtained by completing $X$ at $V$. This is the topologically locally ringed space whose underlying topological space is the same as that of $V$, but whose sheaf of rings is given by taking the ring of functions on $X$ completed at the ideal defining $V$. This is a much better notion of an "infinitesimal neighborhood of $V$"; I usually tell people it is something like an "infinitesimal tubular neighborhood." Formal schemes are described briefly in Hartshorne, but as in Reimundo Heluani's comment, EGA is a much better reference (I think this is EGA $\text{III}_1$, which is a comparatively gentle part of EGA).

In particular, the formal scheme obtained this way, which I'll call $\hat V$, remembers the normal bundle of $V$ in $X$; it is the same as the normal bundle of $V$ in $\hat V$, appropriately defined! In certain very nice cases, it is isomorphic to the formal scheme obtained by completing the total space of the normal bundle of $V$ in $X$ at its zero section, but that is not always the case. See e.g. the comments to this Mathoverflow answer of Jason Starr's.