Please correct me if I am wrong, but I really don't see where we need the proper assumption to get a Galois action.
So let $X/k$ be any scheme, $\overline{k}$ a separable (or algebraic) closure of $k$, and $\mathcal{F}$ be any sheaf on $X$. Write $X_\overline{k}$ for the base change of $X$, $p:X_\overline{k}\rightarrow X$ the induced morphism. Write also $\mathcal{F}_\overline{k}=p^*\mathcal{F}$ for the pullback of $\mathcal{F}$ to $X_\overline{k}$.
For $g\in\mathrm{Gal}(\overline{k}/k)$, we have an induce morphism $g:X_\overline{k}\rightarrow X_\overline{k}$, which induce a pullback $g^*\mathcal{F}_\overline{k}$. I claim that there is a canonical isomorphism $g^*\mathcal{F}_\overline{k}\simeq\mathcal{F}_\overline{k}$. This is because $\mathcal{F}$ comes from $X$. Indeed : $g^*\mathcal{F}_\overline{k}=g^*p^*\mathcal{F}=p^*\mathcal{F}=\mathcal{F}_\overline{k}$.
Hence we have an induced action on cohomology :
$$ H^i(X_\overline{k},\mathcal{F}_\overline{k})\rightarrow H^i(X_\overline{k},g^*\mathcal{F}_\overline{k})\simeq H^i(X_\overline{k},\mathcal{F}_\overline{k})$$
which is the action you are looking for.
Now there is also the approach of Alex Youcis : if $f:X\rightarrow\operatorname{Spec}k$ is the structural morphism, $R^if_*\mathcal{F}$ is a sheaf on $\operatorname{Spec}k$ hence a set equipped with a continuous discrete $\operatorname{Gal}(\overline{k}/k)$-set. Unless I'm mistaken, I don't think we need to add any assumption for the following claim : the underlying $\operatorname{Gal}(\overline{k}/k)$-set is exactly $H^i(X_\overline{k},\mathcal{F}_\overline{k})$ with the above action. (This prove in particular that the above action is continuous).
This follows from the following continuity result
$$\varinjlim_{k'}H^i(X_{k'},\mathcal{F}_{k'})=H^i(X_\overline{k},\mathcal{F}_\overline{k})$$
where $k'$ runs through the finite Galois extensions of $k$ in $\overline{k}$ and the trivial base change $(R^if_*\mathcal{F})_{k'}=R^if_*\mathcal{F}_{k'}$.
EDIT : Let me expand a bit on the last assertion. First, let us show that the stalk of $R^nf_*\mathcal{F}$ at a geometric point $\operatorname{Spec}\overline{k}$ is indeed $H^n(X_\overline{k},\mathcal{F}_\overline{k})$.
By definition, the stalk of $R^nf_*\mathcal{F}$ is $\varinjlim_{k'}R^nf_*\mathcal{F}(k')$ where the limit is taken over all the finite extension of $k'$ inside $\overline{k}$. Now recall that $R^nf_*\mathcal{F}$ is the sheaf associated to the presheaf $k'\mapsto H^n(X_{k'},\mathcal{F}_{k'})$, and since the stalk of a presheaf is the same as its associated sheaf, we get $(R^nf_*\mathcal{F})_\overline{k}=\varinjlim_{k'}H^n(X_{k'},\mathcal{F}_{k'})=H^n(X_\overline{k},\mathcal{F}_\overline{k})$, the last equality is from a limit argument (we should add $X$ quasi-compact and quasi-separated here).
Recall that the equivalence between sheaves on $(\operatorname{Spec}k)_{ét}$ and $\operatorname{Gal}(\overline{k}/k)$-sets is the following : if $\mathcal{F}$ is a sheaf on $\operatorname{Spec}k$, then $\mathcal{F}_\overline{k}=\varinjlim\mathcal{F}(k')$ is a $\operatorname{Gal}(\overline{k}/k)$-set. The action of $\sigma\in\operatorname{Gal}(\overline{k}/k)=\varprojlim\operatorname{Gal}(k'/k)$ on $\mathcal{F}_\overline{k}$ is induced by the compatibles actions on the $\mathcal{F}(k')$ (check that is indeed compatible).
Now you should convinced yourself that this is indeed the pull-back to $\overline{k}$ and that this action is the same as the induced action by functoriality (that is very first one I wrote). Do the same with $R^if_*$ and you will get the compatibility of the two actions in general.
Exercise : Take $k=\mathbb{R}$ and $\mathcal{F}=\mu_4$ the sheaf $A\mapsto\{x\in A, x^4=1\}$ and compute the action of the stalk (there is no trap here).
Just to clarify, the real subtle point here is the following. For a $k$-scheme $X$ let us denote by $|X|$ the underlying topological space. Then, the behind-the-scenes complication here is that $|X\times_k X|\ne |X|\times |X|$. In fact, there is a continuous surjection $|X\times_k X|\to |X|\times |X|$ which is, in general, not an isomorphism (exercise!).
So, if we have a map $\phi:G\times_k G\to G$ and we have $(g,h)\in |G|$ there is no way to make sense of $\phi(g,h)$--one would have to try and lift them to $|G\times_k G|$ and apply $\phi$ there, but this is not well-defined (exercise!).
To convince yourself of why this is not so weird, consider $G=\mathrm{GL}_2$. Note then that we have that $H=\mathrm{SL}_2$ is a closed subscheme of $G$. Let $\eta$ denote the genric point of $G$ and $\eta'$ the generic point of $H$. Then, we have that $k(\eta)=k(x,y,z,w)$ and $k(\eta')=\mathrm{Frac}(k[x,y,z,w]/(xy-zw-1))$. The maps $ \mathrm{Spec}(k(\eta))\hookrightarrow G$ and $\mathrm{Spec}(k(\eta'))\hookrightarrow G$ correspond to the matrices $\begin{pmatrix}x & y\\ z & w\end{pmatrix}$ interpreted in $k(\eta)$ and $k(\eta')$ respectively. How do you 'multiply' those?
The point though is that while $|G\times_k G|\ne |G|\times |G|$ for every $k$-scheme $S$ we have that $(G\times_k G)(S)=G(S)\times G(S)$. Thus, for every $k$-scheme $S$ the map $\phi:(G\times_k G)(S)\to G(S)$ is a map $G(S)\times G(S)\to G(S)$ which is actually the multiplication for a group structure on $G(S)$.
Best Answer
$\newcommand{\Spec}{\mathrm{Spec}}$Let $X$ be a scheme over $k$. Note that $X_{\overline{k}}=X\times_{\mathrm{Spec}(k)}\mathrm{Spec}(\overline{k})$. So, to give a morphism $X_{\overline{k}}\to X_{\overline{k}}$ it suffices to give maps of $k$-schemes $X\to X$ and $\mathrm{Spec}(\overline{k})\to\mathrm{Spec}(\overline{k})$. If $\sigma\in\mathrm{Gal}(\overline{k}/k)$ then declare that the action of $\sigma$ on $X$ is trivial and that the action of $\sigma$ on $\mathrm{Spec}(\overline{k})$ is the one induced from the $k$-algebra map $\overline{k}\to\overline{k}$. The induced map $X_{\overline{k}}\to X_{\overline{k}}$ is the map $\sigma$. Since it's a map of schemes, it's also a map of topological space (read the remark at the end of the post to see a different convention).
If $X=\mathrm{Spec}(A)$ then $X_{\overline{k}}=\mathrm{Spec}(A\times_k \overline{k})$ and $\sigma$ is the map of schemes $X_{\overline{k}}\to X_{\overline{k}}$ corresponding to the ring map $A\otimes_k \overline{k}\to A\otimes_k \overline{k}$ given by $a\otimes \alpha\mapsto a\otimes \sigma(\alpha)$. Covering $X$ by affines with $k$-rational gluing data allows you to bootstrap this concrete understanding of the map to the general case.
How does this relate to the action on $X(\overline{k})$? Note that $$X(\overline{k})=\mathrm{Hom}_k(\mathrm{Spec}(\overline{k}),X)=\mathrm{Hom}_{\mathrm{Spec}(\overline{k})}(\mathrm{Spec}(\overline{k}),X\times_{\mathrm{Spec}(k)}\mathrm{Spec}(\overline{k}))$$
The action on the first presentation of $X(\overline{k})$ is to take $x:\Spec(\overline{k})\to X$ and then $\sigma(x)$ is defined to be the composition
$$\mathrm{Spec}(\overline{k})\xrightarrow{\sigma}\Spec(\overline{k})\xrightarrow{x}X$$
where, again, the first map $\sigma$ is that induced by the ring map $\sigma:\overline{k}\to\overline{k}$. What is the action then on the second presentation of $X(\overline{k})$? It can't clearly be the same idea that for $y:\Spec(\overline{k})\to X\times_{\Spec(k)}\Spec(\overline{k})$ we just precompose by $\sigma$ since that won't be a morphism over $\Spec(\overline{k})$. What's true is that $\sigma(y)$ is the composition
$$\Spec(\overline{k})\xrightarrow{\sigma}\Spec(\overline{k})\xrightarrow{y}X\times_{\Spec(k)}\Spec(\overline{k})\xrightarrow{\sigma^{-1}}X\times_{\Spec(k)}\Spec(\overline{k})\qquad (1)$$
To convince ourselves of this, we need to remind ourselves how the identification
$$\mathrm{Hom}_k(\mathrm{Spec}(\overline{k}),X)=\mathrm{Hom}_{\mathrm{Spec}(\overline{k})}(\mathrm{Spec}(\overline{k}),X\times_{\mathrm{Spec}(k)}\mathrm{Spec}(\overline{k}))$$
works. It takes an $x:\Spec(\overline{k})\to X$ and maps to $y:\Spec(\overline{k})\to X\times_{\Spec(k)}\Spec(\overline{k})$ which, written in coordinates, is the map $(x,\mathrm{id})$. So, we see that $\sigma(y)$ should just be $(x\circ\sigma,1)$. Let's now think about the composition in $(1)$ looks like when written in coordinates. The projection to $X$ is the composition
$$\Spec(\overline{k})\xrightarrow{\sigma}\Spec(\overline{k})\xrightarrow{(x,1)}X\times_{\Spec(k)}\Spec(\overline{k})\xrightarrow{\sigma^{-1}}X\times_{\Spec(k)}\Spec(\overline{k})\xrightarrow{p_1}X$$
Since $p_1\circ\sigma^{-1}$ is just the identity map we see that we get $x\circ\sigma$. What happens we do the second projection? We are looking at the composition
$$\Spec(\overline{k})\xrightarrow{\sigma}\Spec(\overline{k})\xrightarrow{(x,1)}X\times_{\Spec(k)}\Spec(\overline{k})\xrightarrow{\sigma^{-1}}X\times_{\Spec(k)}\Spec(\overline{k})\xrightarrow{p_2}\Spec(\overline{k})$$
which gives us $\sigma^{-1}\circ \sigma=1$. Thus, we see that the composition in $(1)$ is $(x\circ\sigma,1)$ as desired!
Let's do a concrete example. Let's take $X=\mathbb{A}^1_k$. Then, the map $\sigma:X_{\overline{k}}\to X_{\overline{k}}$ is the map corresponding to the ring map $\overline{k}[x]\to \overline{k}[x]$ defined by sending
$$\sum_i a_i x^i\mapsto \sum_i \sigma(a_i)x^i$$
The points of $X_{\overline{k}}$ come in two forms:
It's then clear that $\sigma$ sends $(x-\alpha)$ to $(x-\sigma^{-1}(\alpha))$ (recall that the induced map on $\Spec$ is by pullback! See the remark below) and sends the generic point to itself. That's what it looks like topologically. Let's think about what the action of $\sigma$ looks like on $\overline{k}$-points.
If we take an $\overline{k}$-point corresponding to the map (thinking of $X(\overline{k})$ as $\mathrm{Hom}_k(\Spec(\overline{k}),X)$)
$$x:k[x]\to \overline{k}:p(x)\mapsto p(\alpha)$$
then $\sigma(x)$, on the level of ring maps, is apply $\sigma$ as a poscomposition:
$$\sigma(x):k[x]\to \overline{k}:p(x)\mapsto \sigma(p(\alpha))=p(\sigma(\alpha)$$
where last commutation was because $p(x)\in k[x]$. Let's now think about the same situation in the second presentation $X(\overline{k})=\mathrm{Hom}_{\overline{k}}(\Spec(\overline{k})),X\times_{\Spec(k)}\Spec(\overline{k}))$. Our point $x$ above now corresponds to ring $\overline{k}$-algebra map
$$\overline{k}[x]\to \overline{k}:q(x)\mapsto q(\alpha)$$
If we just postcompose this ring map with $\sigma$ we get
$$\overline{k}[x]\to \overline{k}:q(x)\mapsto \sigma(q(\alpha))$$
which is NOT the desired map sending $q(x)\mapsto q(\sigma(\alpha))$ since $q$ doesn't have rational coefficients. But, if we apply the map $\sigma_X^{-1}$ (where I'm using the subscript $X$ to not confuse it with $\sigma$ on $\overline{k}$) on $X_{\overline{k}}$ this has the effect of applying $\sigma^{-1}$ to the coefficients of $q(x)$. So then, you can see that
$$\sigma(\sigma_X^{-1}(q)(\alpha))=q(\sigma(\alpha))$$
yay!
Hopefully this all makes sense.
NB: Depending on the author one can take the action on $X\times_{\Spec(k)}\Spec(\overline{k})$ to be what I have called $\sigma^{-1}$. This has a couple nice effects:
It's one downside is that ring theoretically it's less natural since then its action on $a\otimes\alpha\in A\otimes_k\overline{k}$ is $a\otimes\sigma^{-1}(\alpha)$. Either convention is OK--nothing changes theory wise--it's just something you have to pay attention/be consistent about.