In general, if there exists a $k$-descent $\xi_0$ of some object $\xi$ over $\overline{k}$, then another $k$-descent is also called a ``twist'' of $\xi_0$. You should then expect the set of twists to be related to $H^1(k,\text{Aut}(\xi))$, where the automorphism group is suitably interpreted. For example, the situation for twists of curves is described in Silverman's Arithmetic of elliptic curves, chapter X.2.
Returning to the situation at hand, a degree $n$ finite etale cover of $\overline{X}$ corresponds to a homomorphism $\overline{\rho} : \pi_1(\overline{X})\rightarrow S_n$, up to conjugation in $S_n$. Its descents to $X$, if they exist, correspond to ways of extending $\overline{\rho}$ to a representation $\rho : \pi_1(X)\rightarrow S_n$, where $\pi_1(\overline{X})$ is embedded inside $\pi_1(X)$ via the base change map $\overline{X}\rightarrow X$.
If we assume that that there exists a descent $f_0$ of your cover $f : Y\rightarrow\overline{X}$, then there is a natural action of $\text{Gal}(k)$ on $\text{Aut}(f)$, and the set of descents should be in bijection with the group cohomology $H^1(\text{Gal}(k),\text{Aut}(f))$, where $\text{Aut}(f)$ is the group of deck transformations of the cover $f$, with the given $\text{Gal}(k)$-action.
I expect the bijection should be given as follows. Let $f_0'$ be another descent of $f$. Then $\overline{f_0'} := f_0'\times_k \overline{k}$ is isomorphic, over $\overline{k}$, to $\overline{f_0} = f$. If $\phi : \overline{f_0'}\rightarrow f$ is an isomorphism (over $k'$), then for any $\sigma\in\text{Gal}(k)$, $\phi^\sigma\circ\phi^{-1}\in\text{Aut}(f)$, and this defines a 1-cocycle $\text{Gal}(k)\rightarrow\text{Aut}(f)$ relative to the above mentioned action of $\text{Gal}(k)$ on $\text{Aut}(f)$. A different choice of isomorphism $\phi$ will give rise to a different cocycle, but the two cocycles give the same class in $H^1(\text{Gal}(k),\text{Aut}(f))$. This should yield a bijection between $H^1(\text{Gal}(k),\text{Aut}(f))$ with the set of descents of $f$.
If the homotopy exact sequence for $\pi_1(X)$ is split -- i.e., $\pi_1(X)\cong\pi_1(\overline{X})\rtimes \text{Gal}(k)$ (e.g., if $X$ admits a $k$-rational point, or $\text{Gal}(k)$ is a projective profinite group, e.g. if $k$ is finite), then we can check this explicitly as follows:
First note that a map $G\rtimes H\rightarrow S_n$ is the same as giving maps $\varphi : G\rightarrow S_n$ and $\psi : H\rightarrow S_n$ such that
$$\varphi(h.g) = \psi(h)\varphi(g)\psi(h)^{-1} \text{ for all $g\in G, h\in H$}. \qquad\qquad(*)$$
Equivalently, if $\alpha : H\rightarrow\text{Aut}(G)$ is the action homomorphism, this says that $\varphi\circ \alpha(h) = \text{inn}(\psi(h))\circ\varphi$ for all $h\in H$. If $\psi' : H\rightarrow S_n$ is another map satisfying this property, then we must have $\text{inn}(\psi'(h)^{-1}\psi(h))\circ\varphi = \varphi$, which is to say $\psi'(h)^{-1}\psi(h)\in C_{S_n}(\text{Im}(\varphi))$. The map $h\mapsto \psi'(h)^{-1}\psi(h)$ is a 1-cocycle $\delta : H\rightarrow C_{S_n}(\text{Im}(\varphi))$, where $h\in H$ acts on $C_{S_n}(\text{Im}(\varphi))$ by conjugation by $\psi(h)$. Note that $\psi(h)$ normalizes $\text{Im}(\varphi)$, and hence also normalizes $C_{S_n}(\text{Im}(\varphi))$, so this makes sense.
In other words, if we take $G = \pi_1(\overline{X})$ and $H = \text{Gal}(k)$, then the set of all descents/twists of $f$ (i.e., extensions of of $\varphi$) are represented by $(\varphi,\delta\psi)$, where $\psi$ is a fixed map satisfying $(*)$ above, and $\delta$ is a 1-cocycle $H\rightarrow C_{S_n}(\text{Im}(\varphi))$.
One should then check that two pairs $(\varphi,\delta\psi)$, $(\varphi,\delta'\psi)$ are conjugate (in $S_n$) if and only if $\delta,\delta,$ represent the same class in $H^1(\text{Gal}(k),C_{S_n}(\text{Im}(\varphi)))$, or in other words, the descents are in bijection with $H^1(\text{Gal}(k),C_{S_n}(\text{Im}(\varphi)))$. I am too lazy to check this carefully but I'm sure it's true (write a comment if it's not!). Noting that $\text{Aut}(f)\cong C_{S_n}(\text{Im}(\varphi))$, we find that this description is compatible with the geometric description given earlier.
In the special case that $\alpha$ is the trivial action, there exists a descent where the homomorphism $\psi$ is zero (and hence $\text{Aut}(f)$ has trivial $\text{Gal}(k)$-action). If moreover $f$ is Galois (i.e. $\varphi$ is the left regular representation), then since the centralizer of the left regular representation is the right regular representation, $\psi(H)$ acts by right multiplication on the Galois group, which can be identified with the image of $\varphi$, and hence every nontrivial $\psi$ will have no fixed points (i.e., $\psi = 0$ is the unique twist which has a fixed point). If moreover the splitting of the homotopy exact sequence comes from a rational point $x$, then this shows that there is a unique descent of $f$ which admits a rational point lying above $x$, and that this descent is totally split above $x$.
