Your attempt at writing things down more explicitly is good. The important thing to realize is that because the equivalence between affine schemes and rings is contravariant, the order of composition changes and swaps left/right actions.
If the authors do not specify whether an action is a left/right action, there is a good chance it is either inferrable from the context, or does not matter. If you can point to specific examples where you think it does matter and it's not clear, these would be good things to ask about as a separate question.
For your explicit example involving actions on $\operatorname{Spec} \Bbb Q[x]$, note that all automorphisms you mention fix $\Bbb Q[x]$ (because they fix $\Bbb Q$) and so they're the identity map. If you instead talk about $\operatorname{Spec} \Bbb Q(i)[x]$, then you do see some movement happening: $(x-i)$ is swapped with $(x+i)$, for instance. One classic fact to learn (mentioned early on in Vakil, for instance) is that if $k\subset K$ is a Galois extension, then $Gal(K/k)$ acts on $\Bbb A^n_K$ and the orbits are precisely the points of $\Bbb A^n_k$. Another good thing to know is that if we have an automorphism $\sigma:k\to k$, then the induced action of $\sigma$ on the $k$-rational points of $\Bbb A^n_k$ is $(a_1,\cdots,a_n)\mapsto (\sigma(a_1),\cdots,\sigma(a_n))$ (proof: just write down the action on the maximal ideal $(x_1-a_1,\cdots,x_n-a_n)$). So these two facts should give you a full understanding of what the Galois action does on affine space.
For subvarieties of affine space, you may need to be a little careful when you define actions. If you want to define an action on $V(I)\subset \Bbb A^n_k$, then you'll need the Galois action to fix $I$. For an example of why this is necessary, think about $V(x-i)\subset \operatorname{Spec} \Bbb Q(i)[x]$. The Galois action here doesn't fix this subvariety and thus does not define an automorphism of it.
As far as group actions on non-affine schemes, things can get a little hairy depending on the specific context. Fortunately, in the case of the Galois action on a scheme $X$ over a field $k$, everything is induced from what happens on $\operatorname{Spec} k$: we define the action of $\sigma \in Gal$ on $X$ to be the map $X\times_k \operatorname{Spec} k\cong X\to X$ induced by taking the fiber product of $X\to \operatorname{Spec} k$ with the automorphism $\sigma:\operatorname{Spec} k\to\operatorname{Spec} k$. If you're interested in studying this action on $X$ via an embedding $X\to Y$, then you need to make sure that the embedding respects the action (this is often described in the literature as "the morphism is an intertwiner" or the like - it means that the morphism commutes with the automorphism).
This brings us to your final point - when we have a morphism of varieties and we want to think about group actions on the source and target, it's usually important to us that the morphism respects the actions: if $f:X\to Y$ is our morphism, we want $g\cdot f(x)=f(g\cdot x)$. Sometimes we don't mean this, though, and if we have an action on $X$, then we can get an action on $\operatorname{Hom}(X,Y)$ by precomposing a map $f:X\to Y$ with an automorphism $\sigma:X\to X$, or if we have an action on $Y$, then we can get an action on $\operatorname{Hom}(X,Y)$ by postcomposing a map $f:X\to Y$ with an automorphism $\sigma:Y\to Y$. So the "action on a morphism" you mention in the comments is just a version of this. (Specifically, saying that $\sigma(F)=F$ is the second version of this - $\sigma$ acts on $Y$, which induces an action on maps as described, if you want to think of it this way. I usually don't, though this isn't my main area and I'm a mathematician, not a cop.)
As for the descent question, the goal is to have $\phi_{\sigma\tau}:(\sigma\tau)V\to V$ factor as $\sigma(\tau V)\stackrel{\sigma\circ\phi_\tau}{\longrightarrow}\sigma V\stackrel{\phi_\sigma}{\longrightarrow} V$, which is just the cocycle condition from (non-Galois) descent.
The diagonal of a Deligne-Mumford stack is not necessarily quasi-separated; take the classifying stack $BG$ of the non-quasi-separated scheme $G = \mathbb{A}^{\infty} \cup_{\mathbb{A}^{\infty} \setminus 0} \mathbb{A}^{\infty}$ viewed as a group scheme over $\mathbb{A}^{\infty} = {\rm Spec} k[x_1, x_2, \ldots, x_n]$.
To solve the exercise, first show that $G_x$ is an etale group algebraic space over $k$. If $k= \bar{k}$, use that any section of the structure morphism $G_x \to {\rm Spec} k$ is an open immersion to give an open covering of $G_x$ by schemes, and then use that any group scheme over a field is separated (see Tag 047L). Over a general field, apply effective descent for separated and locally quasi-finite morphisms to conclude that $G_x$ is a scheme. This shows that the stabilizer of any field-valued point of a Deligne-Mumford stack (w/out any conditions on the diagonal) is a separated etale group scheme.
Best Answer
In general for any $x \in M$, there exists a pointed affine scheme $(U,u)$ and an action $\Gamma_\bar{x}$ on $U$ fixing $u$ and an étale map $[U/\Gamma_\bar{x}] \xrightarrow{f} \mathcal{M}$ such that $f(u) = x$. Then it suffices to compute the $\Gamma_\bar{x}$ on $\mathcal{O}_{U,u}$.
If $\mathcal{M}$ is presented as a global quotient $[V/G]$ where $G$ is reductive, then for every closed point $v \in V$, the Luna Slice Theorem allows one to construct a slice $(U,v)$ and an action of the stabilizer $G_v$ such that $[U/G_v]$ gives us such a chart.
This is the case for $\overline{\mathcal{M}}_{1,1}$ (away from characteristic $2,3$) which can be presented as the quotient $[\mathbb{A}^2/\mathbb{G}_m]$ with the action $g(a,b) = (g^4a, g^6b)$ and universal family $y^2 = x^3 - ax -b$. The point you want is $(1,0)$ which has stabilizer $\mu_4$ the group of fourth roots of unity. A Luna slice is given by $\{(1,t)\}$ with family $y^2 = x^3 - x - t$. The action (as it's induced by the restriction of the $G$ action on $V$) is simply $t \mapsto \zeta_4^6t= \zeta_4^2t$ and the completed local ring is $k[[t]]$ with this action of $\mu_4$.
It's not hard to see using the same method that at the point classifying $y^2 = x^3 - 1$, ${(t,1)}$ us a Luna slice and $t \mapsto \zeta_6^4t$, and at any other point the automorphism group $\mu_2$ acts trivially.
If the stack isn't given to you as a global quotient I don't see any general way to answer the question. You just need to compute a versal deformation and its automorphisms. One observation that helps when $\mathcal{M}$ is smooth though is that in this case we can arrange it so that there is an étale map $[U/\Gamma_\bar{x}] \to [T_x/\Gamma_\bar{x}]$ and so the question reduces to understanding the linear action of $\Gamma_\bar{x}$ on the tangent space. For example for the moduli of curves, we need to understand how $\mathrm{Aut}(C)$ acts on $H^1(C,T_C)$ which is not much more explicit but at least reduces the question to a cohomology computation.