Question 1: “Fix an algebraically closed field k. The affine varieties over k and their morphisms, by definition, form a full subcategory of the category LRS of locally ringed spaces. Does it make sense to think of the full subcategory of LRS whose objects are the affine varieties over any algebraically closed field? That is, to think of morphisms of LRS between affine varieties over different fields?”
Answer: Yes, this makes sense. If $k \rightarrow L$ is an embedding of fields and $A$ a $k$-algebra, $B$ a $L$ algebra, it makes sense to speak of maps
$$\phi: X:=\mathrm{Spec}(B) \rightarrow S:=\mathrm{Spec}(A).$$
Whenever you have a point $x \in X$ with image $f(x):=s \in S$
you get an induced extension of residue fields $\kappa(s) \subseteq \kappa(x)$,
hence for this to make sense, your fields must have the same characteristic.
Note that if $k,L$ are field with $char(k) \neq char(L)$ and if $X$ is a scheme over $k$, $Y$ a scheme over $L$, there are no morphisms $f: X \rightarrow Y$. Hence $X$ cannot be a scheme over $k$ and $L$. Hence if $Aff(k)$ (resp $Aff(L)$) are the "categories of affine schemes over $k$" (resp $L$) there is no object $X \in Ob(Aff(k))$ and $Ob(Aff(L))$. The two categories have no "objects in common".
Question 2: “Further, an algebraic variety is a LRS with an open covering of affine varieties (satisfying some extra conditions to prevent the case of a general prevariety), do the affine varieties of the covering need to be over the same field?”
If a variety/scheme $X$ is defined over a field $k$ it follows any open subscheme (or subvariety) $U:=\mathrm{Spec}(A) \subseteq X$ is defined over $k$.
Let $k'$ be another field; if $X$ and $Y$ are affine varieties over $k$ and $k'$ respectively, how to define an arbitrary morphism $X\to Y$ in this setting?
This is a fine question but it doesn't really have anything to do with (locally) ringed spaces, and makes perfect sense using completely algebraic foundations, e.g. defining $\text{Aff}_k$ to be the opposite of the category of finitely generated integral domains over $k$ or similar.
$X$ and $Y$ are not objects in the same category and so there is no obvious way to talk about morphisms between them. However, if we have the additional data of a morphism $f : k \to k'$, then there is a base change functor $(-)_{k'} : \text{Aff}_k \to \text{Aff}_{k'}$ which in terms of $k$-algebras is given by the tensor product $(-) \otimes_k k'$, and then given $X \in \text{Aff}_k$ and $Y \in \text{Aff}_{k'}$ we can meaningfully ask for a morphism $X_{k'} \to Y$ in $\text{Aff}_{k'}$. More generally if we have the additional data of a third field $k''$ together with two morphisms $k \to k'', k' \to k''$ then we can base change both $X$ and $Y$ to $k''$.
Edit: There is a mild subtlety here; we actually need to work with affine schemes. The problem is that the base change of an affine variety can fail to be an affine variety, because the base change of an integral domain can fail to be an integral domain. The next paragraph gives a simple example.
In general the tensor product of fields is not a field, but maybe if $k$, $k'$ have same characteristic $p$, then $k\otimes_{\mathbb{Z}_p}k'$ is always a field?
No, e.g. the tensor product of $\mathbb{F}_{p^n}$ with itself is never a field if $n \ge 2$. It is a nice exercise to show more generally that if $L$ is a finite Galois extension of $K$ with Galois group $G$ then $L \otimes_K L \cong L^G$.
However those in $\mathrm{Aff}_k$ are not the only affine varieties in $\mathrm{RS}_k$; e.g., also affine varieties over $k'\subset k$ are objects of $\mathrm{RS}_k$. Is there some convention on this?
Answering this question involves clarifying a subtle conceptual point. It is not the case, strictly speaking, that affine varieties over $k' \subset k$ are objects of $\text{RS}_k$. What is true is that there is a functor $\text{Aff}_{k'} \to \text{RS}_k$, given by base change as above; however, this functor is not full, because base changing to $k$ causes more morphisms to exist. So it's misleading to think of the image of this functor as consisting of affine varieties over $k'$, because the morphisms are different (and so e.g. there can be varieties which are not isomorphic over $k'$ but which become isomorphic when base changed to $k$). What it consists of is affine varieties $X$ over $k$ which happen to be the base change of an affine variety $X'$ over $k'$; a choice of such an affine variety is called a $k'$-form of $X$, and it's a delicate and subtle question when they exist and how many there are when they do, involving what is called Galois descent.
A simple example here is to take $k' = \mathbb{R}, k = \mathbb{C}$ and consider the varieties
$$V = \{ x^2 + y^2 = -1 \}, W = \{ x^2 + y^2 = 1 \}$$
over $\mathbb{R}$. These varieties are not isomorphic over $\mathbb{R}$ because $V$ has no real points but $W$ does. However, after complexification (base change to $\mathbb{C}$) they become isomorphic because we can scale $x$ and $y$ by $i$. So $V$ and $W$ are two non-isomorphic real forms of the same complex variety, which turns out to be the punctured affine line $\mathbb{A}^1 \setminus \{ 0 \}$ (this is a nice exercise; the point is that over $\mathbb{C}$ we can write $x^2 + y^2 = (x + iy)(x - iy)$).
Best Answer
No, as the following example proves.
Example. Let $\mathbb{K}=\overline{\mathbb{Z}_p}$ be the algebraic closure of the field of integer numbers modulo $p\in\mathbb{P}$, and let $X=Y=\mathbb{A}^1_{\mathbb{K}}$ as locally ringed space; considering the following maps of $X$ in itself: \begin{gather} O:x\in X\to 0\in X,\\ F:x\in X\to x^p-x\in X; \end{gather} one has that as maps $O=F$, but as regular maps $O\neq F$ because $O^{*}$ is the zero endomorphism of $\mathbb{K}[t]$ and $F^{*}$ is the Frobenius endomorphism of $\mathbb{K}[t]$. $\triangle$
And if one fixes a continuous map $f:(X,\mathcal{O}_X)\to(Y,\mathcal{O}_Y)$ then the pull-back $f^{*}:\mathcal{O}_Y\to f_{*}\mathcal{O}_X$ is well-defined ever; and if $f^{*}$ is a morphism of sheaves then $(f,f^{*})$ is a morphism of locally ringed spaces.