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).
The main theorem to mention is:
Theorem. (Belyi) Let $X$ be a smooth projective curve over $\mathbb C$. Then $X$ is defined over $\bar {\mathbb Q}$ if and only there exists a map $X \to \mathbb P^1_{\mathbb C}$ ramified over at most three points.
Remark. The 'only if' part is due to Belyi in 1979, and is basically an algorithm. The 'if' part was known before that (I believe it was proven by Weil in 1956). See this paper for a modern account of the 'if' part, and some background.
Note that three points on $\mathbb P^1$ are in general position, i.e. for any ordered set $(x,y,z)$ of distinct [closed] points, there exists an automorphism of $\mathbb P^1$ mapping $(x,y,z)$ to $(0,1,\infty)$. This is just linear algebra: we can choose coordinates so that $x$, $y$, and $z$ correspond to $[0:1]$, $[1:1]$, and $[1:0]$ respectively. Thus, if $X$ is a curve defined over a number field $K$, then we get a finite étale map
$$U \to \mathbb P^1_K\setminus\{0,1,\infty\}$$
of some open $U \subseteq X$. This suggests that the study of $\pi_1^{\operatorname{\acute et}}(\mathbb P^1_{\mathbb Q} \setminus\{0,1,\infty\})$ could be interesting for number theoretic purposes.
Indeed, the 'fibration'
$$\begin{array}{ccc}\mathbb P^1_{\bar{\mathbb Q}}\setminus\{0,1,\infty\} & \to & \mathbb P^1_{\mathbb Q} \setminus\{0,1,\infty\} \\ & & \downarrow \\ & & \operatorname{Spec} \mathbb Q \end{array}$$
induces a short exact sequence
$$1 \to \pi_1^{\operatorname{\acute et}}(\mathbb P^1_{\bar{\mathbb Q}}\setminus\{0,1,\infty\}) \to \pi_1^{\operatorname{\acute et}}(\mathbb P^1_{\mathbb Q}\setminus\{0,1,\infty\}) \to \pi_1^{\operatorname{\acute et}}(\operatorname{Spec} \mathbb Q) \to 1.$$
The first group is the free profinite group $\hat F_2$ on two generators, since a change of algebraically closed field does not alter the fundamental group (and over the complexes we get the profinite completion of the topological fundamental group). The last group, of course, is just $\operatorname{Gal}(\bar{\mathbb Q}/\mathbb Q)$.
Thus, this in turn defines a map
$$\operatorname{Gal}(\bar{\mathbb Q}/\mathbb Q) \to \operatorname{Aut}(\hat F_2) \twoheadrightarrow \operatorname{Out}(\hat F_2).$$
Corollary (of Belyi's theorem). The map $\operatorname{Gal}(\bar{\mathbb Q}/\mathbb Q) \to \operatorname{Out}(\hat F_2)$ is injective.
See Szamuely's Galois groups and fundamental groups, Theorem 4.7.7. This means that we can view $\operatorname{Gal}(\bar{\mathbb Q}/\mathbb Q)$ as a subgroup of an object coming from topology. If we could understand what the image of the above injection is, we would solve all of number theory. This is the motivation for Grothendieck's study of dessins d'enfants.
There are also applications to the inverse Galois problem; see for example [loc. cit.], section 4.8.
See this post for some further applications and alternative perspectives on Belyi's theorem.
Best Answer
As Alex Wertheim has mentioned in the comments, the key idea here is twisted forms. The (informal) general idea is the following: suppose we are given a category $\mathcal C_K$ of structures defined over a field $K$ (for example, $\mathcal C_K$ may be affine $K$-varieties, projective $K$-varieties, $K$-algebras, etc.). Let $L/K$ be a Galois extension with Galois group $G$, and let $\mathcal C_L$ be the corresponding category of structures defined over $L$. Then $G$ acts on each object of $L$ by automorphisms.
The main theorem is the following. Given an object $X$ of $\mathcal C_K$, we can obtain its base change $X_L$, an object of $\mathcal C_L$. Then there will be a bijection $$\{K\text{-isomorphism classes of }\mathcal C_K\text{ objects isomorphic to }X_L\text{ after base change}\}\longleftrightarrow H^1(G,\text{Aut}_{\mathcal C_L}(X_L))$$ where $\text{Aut}_{\mathcal C_L}(X_L)$ has the $G$-conjugation action induced by the action of $G$ on $X_L$. The objects on the left hand side of this correspondence are the so-called "twisted forms." This correspondence probably easiest to see in the case of affine varieties, or equivalently, (reduced f.g.) commutative $K$-algebras. Given an affine $K$-variety $X$ with coordinate ring $K[X]$ and its base change $X_L$ with coordinate ring $L[X]$, a $1$-cocycle $H^1(G,\text{Aut}_L(L[X]))$ canonically defines a semilinear action on $L[X]$. The theory of faithfully flat descent says that the category of $L$-algebras with semilinear $G$-actions is equivalent to the category of $K$-algebras, and so this canonically gives a $K$-algebra which becomes isomorphic to $L[X]$ after base change. A good reference for this fact about twisted forms is Springer's book on algebraic groups.
To explain the correspondence between conics and cocycles, we apply this to the category of projective varieties over $K$ and $X = \mathbb P^1_K$. The left-hand side of our correspondence is then simply the $K$-isomorphism classes of smooth curves of genus $0$, since we are base changing to an algebraically closed field. The right hand side of the correspondence is $H^1(G,\text{Aut}_{\bar K}(\mathbb P^1_{\bar K})) = H^1(G, \text{PGL}_2(\bar K))$. Tracing through the bijection explicitly will tell you which conic corresponds to which cocycle.
As an example, let's consider the simple case of $\mathbb C/\mathbb R$. Then $\mathbb P^1_{\mathbb R}$ embeds in $\mathbb P^2_{\mathbb R}$ as the conic $C : y^2-xz=0$ via $[s:t]\mapsto [s^2:st:t^2]$. Now one checks that there are only two cocycles in $H^1(G,\text{PGL}_2(\mathbb C))$, namely the trivial cocycle and the one which sends complex conjugation to $\left(\begin{smallmatrix} 0 & -1 \\ 1 & 0 \end{smallmatrix}\right)$. This is the automorphism of $\mathbb P^1$ given by $[s:t]\mapsto [-t:s]$ which corresponds to the automorphism of $C$ given by $[x:y:z]\mapsto [z:-y:x]$. To find the conic in $\mathbb P^2_{\mathbb R}$ corresponding to this cocycle, we need to find a conic for which complex conjugation interchanges $x$ and $z$ and negates $y$. So we make the change of variables $x = x'+iz'$, $y=iy'$, $z=x'-iz'$. Then complex conjugation interchanges $x$ and $z$ while negating $y$. The equation of the resulting conic is $$0 = xz-y^2 = x'^2+y'^2+z'^2.$$ So the conic $C' : x'^2 + y'^2 + z'^2 = 0$ is the desired conic, which is an instance of the sort of conic you mention in your question.
As Alex also mentioned, this is all tied up with quaternion algebras, but I won't get too far into that here. The reason is basically that $\text{PGL}_2(\bar K)$ is also the automorphism group of the matrix algebra $M_2(\bar K)$. So 4-dimensional central simple algebras defined over $K$ are also classified by $H^1(G,\text{PGL}_2(\bar K))$. A conic $ax^2+by^2+cz^2$ will correspond to a quaternion algebra easily expressible in terms of the $a,b,c$.