This is part of an exercise (2.14) in Qing Liu's book on Algebraic Geometry. I'm going to be verbose to see if there's something fundamental I'm missing. Actual questions will be numbered.
Let $G$ act on a ringed topological space $(X, \mathcal{O}_X)$, ie we have a group morphism $\phi\colon G \to \operatorname{Aut}_{LRS}(X)$.
The full exercise wants to finally show that the quotient space exists in the category of ringed topological spaces. It is the object defined by the property of being a ringed topological space $(Y, \mathcal{O}_Y)$ together with a quotient morphism $p \colon X \to Y$ that is universal among all maps that satisfy $p \circ \sigma = p$ for all $\sigma \in G$ ($G$-invariance) and that any morphism $f \colon X \to Z$ that is also $G$-invariant factors through $p$ uniquely by $f = \tilde{f} \circ p $ for some $\tilde{f} \colon Y \to Z$.
First I had trouble even with the unwinding of definitions of how the group acts on a locally ringed space. I assume that this means that we can identify the group action on the (topological) set as $(\phi(\sigma), \phi(\sigma)^\#): (X, \mathcal{O}_X) \to (X, \mathcal{O}_X)$. With abuse of notation the topological morphism as $\sigma \colon x \mapsto x\sigma := \phi(\sigma)(x)$ and the map of sheafs is $\sigma^\# \colon \mathcal{O}_X \to \sigma_* \mathcal{O}_X$ or explicitly $\sigma ^\# \colon \mathcal{O}_X(U) \to \mathcal{O}_X(\sigma^{-1}(U))$. Is this correct?
I also assume that the $G$-invariance $p \circ \sigma = p$ can be translated into $p(x \sigma) = p(x)$ as topological spaces and $(p \circ \sigma)^\# = (p_* \sigma^\#) \circ p^\#$ as maps of sheafs so that $\mathcal{O}_X(p^{-1}(U)) = \mathcal{O}_X(\sigma^{-1}(p^{-1}(U)))$. Is this correct?
The quotient is then constructed by $Y = X/G$ as quotient set, and $p \colon X \to Y$ the projection. The topology is given by the quotient topology giving $V \subset Y$ is open if and only if $p^{-1}(U)$ is open in $X$. I've shown that $p$ is an open morphism of topologies.
- Part of the exercise wants one to show how $G$ acts on the ringed space $p^{-1}(V)$ (do they mean $(W, \mathcal{O}_{W})$ for $W = p^{-1}(V)$?) explicitly and that, in particular, $G$ acts on the ring $\mathcal{O}_X(p^{-1}(V))$. This feels very elementary but I can't really see how the group morphism $\phi$ gives a group morphism $G \to \operatorname{Aut}_{LRS}(p^{-1}(V))$ and let alone the action on the ring.
Corrected question 2. with comment from @Mindlack.