The following seems to give a reasonable affirmative answer which avoids computing the regular functions, and replaces condition (2) (whose statement I don't understand:  why is $f$ mentioned there?) with the more natural condition that the subset $\Sigma := X(\overline{k})$ in (1) is stable under the action of the Galois group on $\overline{k}^n$.

Let's be cleaner by working more generally over an arbitrary (not necessarily perfect) field $k$ and with geometrically reduced closed subschemes $X$ in a fixed separated $k$-scheme $Y$ locally of finite type.  (Note: now affine schemes are gone; can take $Y$ to be an affine space if you wish, but this is irrelevant.)  The 
${\rm{Gal}}(k_s/k)$-stable set $X(k_s)$ in $Y(k_s)$ recovers $X$ by the following 2-step procedure.  First, if $A$ is a finitely generated $k$-algebra then $X(A)$ is the set of points $a \in Y(A)$ such that $a(x) = 0$ for all $x \in \Sigma$.
In general, for any $k$-algebra $L$ the subset 
$X(L)$ in $Y(L)$ is the directed union of the sets 
$X(A)$ where $A$ varies through all $k$-subalgebras $A \subseteq L$ of finite type over $k$. Voila, no coordinate rings.  (Remember: we didn't require $Y$ to be affine.) 

Of course, one can apply this process to any ${\rm{Gal}}(k_s/k)$-stable subset $\Sigma$ of $Y(k_s)$, and what is the meaning of the resulting functor?  It represents the Galois descent $X$ of the Zariski closure in $Y_{k_s}$ of $\Sigma$.  In general $X(k_s)$ may be larger than $\Sigma$, but nonetheless $\Sigma$ is Zariski-dense in $X_{k_s}$.  This is perfectly interesting in practice, regardless of whether or not $\Sigma$ is equal to $X_{k_s}$, since it is what underlies the construction of derived groups, commutator subgroups, images, orbits, and related things in the theory of linear algebraic groups over a general field.  For example, the $k$-group ${\rm{PGL}}_n$ is its own derived group in the sense of algebraic groups, but the commutator subgroup of ${\rm{PGL}}_n(k_s)$ is a proper subgroup whenever $k$ is imperfect and ${\rm{char}}(k)|n$. 

To give a nifty application, suppose one begins with an arbitrary closed subscheme $X'$ in $Y$ (such as $X' = Y$!), then forms the ${\rm{Gal}}(k_s/k)$-stable set $X'(k_s)$ (which could well be empty, or somehow really tiny), and then applies the above procedure to get a geometrically reduced closed subscheme $X$ in $X'$.  What is it?  It is the maximal geometrically reduced closed subscheme of $X'$, and one can check its formation is compatible with products (as well as separable extensions $K/k$, such as completions $k_v/k$ for a global field $k$).  If $k$ is perfect then $X = X'_{\rm{red}}$, so this is more interesting when $k$ is imperfect.  It is especially interesting in the special case when $X'$ is equipped with a structure of $k$-group scheme.  Then $X$ is its maximal smooth closed $k$-subgroup, since geometrically integral $k$-groups locally of finite type are smooth.  So what?  If one is faced with the task of studying the Tate-Shararevich set for such an $X'$ (e.g., maybe $X'$ is a nasty automorphism scheme of something nice) then all that really intervenes is $X$ since it captures all of the local points, so for some purposes we can replace the possibly bad $X'$ with the smooth $X$. (This trick is used in the proof of finiteness of Tate-Shafarevich sets for arbitrary affine groups of finite type over global function fields.)  But beware: if the $k$-group $X'$ is connected (and $k$ is imperfect) then $X$ may be disconnected and have much smaller dimension; see Remark C.4.2 in the book "Pseudo-reductive groups" for an example.