Let $K$ be a compact Lie group with $\mathcal k=\text{Lie}(K)$, and choose one vector $v\in \mathcal k$.
Define \begin{equation} \mathcal k_v=\{X\in \mathcal k\colon[X,v]=0 \} \end{equation} then the centre of $\mathcal k_v$ is \begin{equation} \mathcal z_v=\{u\in\mathcal k_v\colon [u,X]=0,\;\forall X\in\mathcal k_v\} \end{equation} and the centralizer of $\mathcal k_v$ in $\mathcal k$ is \begin{equation} Z_v=\{u\in\mathcal k\colon [u,X]=0,\;\forall X\in\mathcal k_v\} \end{equation} then can we recover $\mathcal k_v$ from $\mathcal z_v$ or $Z_v$ by the proving the following: \begin{align} \mathcal k_v\overset{?}=\{X\in \mathcal k\colon [X,u]=0,\;\forall u\in\mathcal z_v \}\\ \mathcal k_v\overset{?}=\{X\in \mathcal k\colon [X,u]=0,\;\forall u\in Z_v \} \end{align}
Thanks in advance.
Not really an answer, and steps 1 and 2 below need double checking, but here's a train of thoughts:
Since this is not true in general, a proof would have to use some property of compact (or as I would call them, anisotropic) Lie algebras. Assuming we talk about semisimple Lie algebras here (well, the abelian case would be trivial), I would bet on the fact that "all nonzero elements of a compact Lie algebra are semisimple" as being crucial.
Namely, for all nonzero $v$, $k_v$ as centralisers of semisimple elements should be Levi subalgebras (step 1), and in the compact setting, the only Levi subalgebras should be maximal tori i.e. Cartan subalgebras (step 2). These are self-centralising and abelian, in particular $k_v=Z_v=z_v$ and your last two relations hold.