(skip to the part about theorem 4.8 if you know things about valuation theory on function fields and local uniformization).
Context:
For rings $R$, denote the quotient field by QF$(R)$. For valuations $v$, let $R_v$ denote the valuation ring, $M_v$ its maximal ideal, $D_v:= R_v/M_v$ the residue field, and $S_v$ the value group. If you are not familiar with quadratic transforms, you can probably ignore that part (I think my question is a basic valuation theoretic one).
Definition 3.10 ([Ramification Theoretic Methods in Algebraic Geometry - Shreeram Abhyanker], or briefly, [A]). If $(R,M)$ is a local domain so that there is a finite type algebra $A$ over a field $k$ s.t. $R = A_p$, where $p = M\cap A$, then we say that $R$ is $\textit{algebraic with ground field $k$}$ and we let $\text{k-rank}(R):= \text{tr.deg}_k R/M$. We have $\text{k-rank}(R) + \dim(R) = \text{tr.deg}_k \text{QF}(R)$.
Theorem 4.8 [A]. Let $L/k$ be a two-dimensional algebraic function field, and $K$ a finite algebraic extension of $L$. Let $v$ be a zero dimensional valuation of $K$ over $k$ (this means that $v\mid_k=0$ and tr.deg$_k D_v=0$). Let $(R,M)$ be a regular algebraic local domain with quotient field $K$ and ground field $k$ s.t.$^1$ $\dim(R) = 2$. Then for some quadratic transform $R^*$ of $R$ along $v$ there exists an algebraic local domain $S^*$ in $L$ with ground field $k$ lying below $R^*$.
(1: he also says: "k-rank$(R) = 2$" but I am fairly certain that is a typo, otherwise 3.10 together with our hypothesis would yield: $4 = 2 + 2 = \text{k-rank}(R) + \dim(R) = \text{tr.deg}_k K = \text{tr.deg}_k L + \text{tr.deg}_L K = 2 + 0 = 2$).
(I am not copying the proof here because I will state (have stated) below (above, resp.) everything that I think has a chance of being relevant to my question).
Question: Relatively early in the proof, Abhyanker uses the supposed fact of $v$ being a rank one valuation, but this is really not clear to me after a good several hours of contemplating.
Notes (what I tried so far): [A] has a characterization of valuations on a two dimensional function field (of which v is: since tr.deg$K/k = 2$), and by this characterization (and using some comments from Dale Cutkosky's $\textit{Resolution of Singularities}$ book (call this [R]) section 8.1) either we get what we want ($v$ has rank one, i.e. $S_v\leq \mathbb{R}$ as ordered abelian groups) or $S_v = (\mathbb{Z}^2)_{\text{lex}}$, but I really cant seem to find an issue with that being the case (I tried using 8.1.2 in [R], noting that, in the latter case, $v$ would be a zero dimensional valuation of rank two).