Here is a paragraph from a paper of Lang (page 385):
Now, I have trouble understanding what does the red underlined line actually mean. What is actually this ring of vectors and why is $T:=F^{\text{ur}}$ the quotient field of that. And are all the entries $\xi_i$ of the vector from the same $E$ or each one is from some arbitrary such $\mathfrak{k}\subset E\subset \bar{\mathfrak{k}}$ depending on $i$?
Here are my thoughts of what it could mean. We know $F^{\text{ur}}$ has residue field $\bar{\mathfrak{k}}$, and thus the completion $\widehat{F^{\text{ur}}}$ of $F^{\text{ur}}$ is complete discretely valued with algebraically closed residue field. It is well-known then (Serre Local fields chapter 2 sections 4 and 5), that then either $\widehat{F^{\text{ur}}}$ is isomorphic $\bar{\mathfrak{k}}((t))$ or to the quotient field of the ring of Witt vectors $W(\bar{\mathfrak{k}})$.
Now indeed in each of these two cases, we can regard the elements of $\widehat{F^{\text{ur}}}$ (in particular those of $F^{\text{ur}}$) as vectors with elements in $\bar{\mathfrak{k}}$. I am not sure that this is what the author means, and could use someone's help.
This is probably going to be too long for a comment.
As is all too often true of the works of the Great and Blameless Serge Lang, this paragraph is extremely poorly written. The truth of the matter is that if $t\in T$, then there is a particular finite extension $E$ of the residue field such that $t$ is representable as an element of the Witt vectors over $E$.
If you allow the entries of your Witt vector to be in differing finite extensions $E_i$ of the residue field, then you get an element of the completion of the maximal unramified.
If this doesn’t answer your question, I’ll do what I can to expand the above.