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There may be some technical issues with the question, but hopefully what I mean is clear...

Let $k$ be a number field (or maybe any finitely generated field over $\mathbb{Q}$ of characteristic 0)

Let $k(\!(t)\!)$ be the field of Laurent series in $t$ with coefficients in $k$, and let $\Omega$ denote an algebraic closure of $k(\!(t)\!)$, and let $\overline{k}$ denote the algebraic closure of $k$ inside $\Omega$.

There is a natural map $$\rho : \mathrm{Gal}(\Omega/k(\!(t)\!))\rightarrow \mathrm{Gal}(\overline{k}/k)$$ given by restriction.

If $\{t_n\}_{n\ge 1}\subset \Omega$ satisfies $t_1 = t$, $t_n^d = t_{n/d}$ for all $d\mid n$, then we say that it is a compatible system of roots of $t$.

Given a compatible system of roots $\{t_n\}$, and a filtration $\overline{k} = \bigcup L$ by finite extensions of $k$, the tensor product $\left(\varinjlim_L L(\!(t)\!)\right)\otimes_{k(\!(t)\!)} \left(\varinjlim_n k(\!(t)\!)(t_n)\right)$ is a field isomorphic to $\Omega$, and thus we obtain an section of the map $\rho$ by having $\sigma\in \mathrm{Gal}(\overline{k}/k)$ act on the tensor product in the obvious way in the first factor, and trivially on the second factor. In particular, this action stabilizes the compatible system of roots $\{t_n\}$.

Does every section of $\rho$ stabilize some compatible system of roots $\{t_n\}$?

Note that the kernel of $\rho$ is the subgroup $\mathrm{Gal}(\Omega/E)$ where $E = \varinjlim_L L(\!(t)\!)$ as before, and the Galois group is isomorphic to $\widehat{\mathbb{Z}}$. The group $\mathrm{Gal}(\overline{k}/k)$ acts on $\mathrm{Gal}(\Omega/E) \cong \widehat{\mathbb{Z}}$ via the cyclotomic character, and if $k$ is finitely generated over $\mathbb{Q}$ then the only element of $\widehat{\mathbb{Z}}$ fixed by all of $\mathrm{Gal}(\Omega/E)$ is the identity. This implies that if a section of $\rho$ comes from a compatible system, then it must be unique.

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$\newcommand{\Gal}{\mathrm{Gal}}\newcommand{\Z}{\mathbb{Z}}$Fix a compatible system $(t_n)$ of roots of $t$. It provides us with a section of $\rho$ thus giving an isomorphism between $\Gal(\overline{k((t))}/k((t)))$ and the semi-direct product $\Gal(\overline{k}/k)\ltimes \hat{\Z}(1)$ where $\hat{\Z}(1)$ denotes $\lim\limits_{\leftarrow}\mu_n(\bar{k})$ as a $\Gal(\bar{k}/k)$-module. An element $(g,m)$ acts on $\overline{k((t))}$ as follows: in terms of your decomposition $$\overline{k((t))}=\left(\varinjlim_L L(\!(t)\!)\right)\otimes_{k(\!(t)\!)} \left(\varinjlim_n k(\!(t)\!)(t_n)\right)$$ it acts on the first factor as prescrivbed by $g$ and, on the second factor sends $t_n$ to $m_nt_n$ where $m_n$ is the projection of $m$ to $\mu_n(\bar{k})$.

If $(t'_n)$ is another compatible system of roots of $t$, it provides another section $\Gal(\bar{k}/k)\to \Gal(\overline{k((t))}/k((t)))$ which is characterized by the fact that it preserves all the elements $t'_n$. In terms of the semi-direct product, this section then looks like $g\mapsto (g,g(m)-m)$ where $m\in\hat{\Z}(1)$ is the element whose components $m_n\in\mu_n(\bar{k})$ are given by $t_n^{-1}t'_n$

In general, if $G\ltimes M$ is the semidirect product of a profinite group $G$ with a continuous module $M$, a giving a continuous section $s:G\to G\ltimes M$ is equivalent to giving a map $f:G\to M$ such that $f(g_1g_2)=g_1f(g_2)+f(g_1)$.

Coming back to our situation we see that section are in bijection with continuous 1-cocycles of the group $\Gal(\bar{k}/k)$ with coefficients in $\hat{\Z}(1)$ and a section comes from a system of roots if and only if this cocycle is a coboundary. In other words, we get

Lemma. Every continuous section comes from a compatible system of roots of $t$ if and only if $H^1_{cont}(\Gal(\bar{k}/k),\hat{\Z}(1))=0$.

In your setting, this group is never zero, for example, from the exact sequence $0\to\hat{\Z}(1)\xrightarrow{n}\hat{\Z}(1)\to \mu_n\to 0$ we see that $n$-torsion in this cohomology group is given by $$H^1_{cont}(\Gal(\bar{k}/k),\hat{\Z}(1))[n]=\mathrm{coker}(\hat{\Z}(1)^{\Gal(\bar{k}/k)}\to \mu_n(k))$$ Since the order of roots of unity contained in $k$ is bounded, $\hat{\Z}(1)^{\Gal(\bar{k}/k)}$ is zero, but, for instance, $\mu_2(k)$ is equal to $\Z/2$ for every $k$.

So, there are always other sections but you can control them by this Galois cohomology group.

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