The construction you describe when $\mathscr{C}$ consists of all closed sets of $X$ is known as the Wallman compactification of $X$. I'll denote if $\omega(X)$. It is due to Wallman; Lattices and topological spaces, Ann. Math. 39 (1938) 112-126.
Of course some sort of techincal assumption is required.
Let $X$ be a $T_1$ space. Then $\omega(X)$ is a compact $T_1$ space containing $X$ as a dense subspace. Moreover it has the property that every continuous map $X\rightarrow K$ into a compact Hausdorff space $K$ extends over $\omega(X)$. The space $\omega(X)$ is Hausdorff if and only if $X$ is normal, and in this case $\omega(X)\cong\beta(X)$.
Shanin later generalised Wallman's construction; On special extensions of topological spaces, Dokl. SSSR 38 (1943) 6-9, On separation in topological spaces, Dokl. SSSR 38 (1943) 110-113, On the theory of bicompact extensions of topological spaces, Dokl. SSSR 38 (1943) 154-156. The compactifications that Shanin constructed allowed for the ultrafilters to come from more general lattices $\mathscr{L}$ of closed subsets of $X$. Of course at the expense of added assumptions: $\mathscr{L}$ is required to be a so-called $T_1$-base for the closed subsets of $X$. Denote by $\omega(X;\mathscr{L})$ the Wallman-Shanin compactification built using the $T_1$-base $\mathscr{L}$.
Here are some examples to convince you that these compactifications are interesting.
- $X$ is locally compact $T_2$ and $\mathscr{L}$ consists of all $(i)$ compact subsets of $X$, and $(ii)$ all closed subsets $A\subseteq X$ for which there is a compact $K\subseteq X$ with $A\cup K=X$. Then $\omega(X;\mathscr{L})$ is the Alexandroff compactification of $X$.
- $X$ is Tychonoff and $\mathscr{L}=\mathscr{Z}(X)$ is the collection of zero sets. Then $\omega(X;\mathscr{L})\cong\beta(X)$, as you have recognised.
- $X$ is rim-compact $T_2$ and $\mathscr{L}$ is the set of all finite intersections of regularly closed sets with compact boundaries. Then $\omega(X;\mathscr{L})=\mathfrak{f}(X)$ is the Freudenthal compactification of $X$.
The answer to both of these questions is Yes. And this result can generalize to point-free topology and I consider this result to be more natural in the context of point-free topology.
The following observations were produced earlier by Mehmet Onat, so let me paraphrase those arguments.
Observation: If $U$ is an open subset of $X$ and $Y$ is dense in $X$, then
$\text{Cl}_X(U\cap Y)=\text{Cl}_X(U)$.
Proof: Clearly $\text{Cl}_X(U\cap Y)\subseteq\text{Cl}_X(U)$. For the converse direction, suppose that $a\in\text{Cl}_X(U)$. Then whenever $O$ is an open subset of $X$ that contains $a$, we have $O\cap U\neq\emptyset$. However, since $Y$ is dense in $X$, we also know that $O\cap U\cap Y\neq\emptyset$. Therefore, $a\in\text{Cl}_X(U\cap Y)$, so we may conclude the converse $\text{Cl}_X(U)\subseteq \text{Cl}_X(U\cap Y).$ $\square$
Claim: Suppose that $Y$ is dense in $X$. If $C\in R(Y)$, then $\text{Cl}_X(C)\in R(X)$.
Proof: If $C\in R(Y)$, then there is some open subset $U\subseteq X$ where
$$C=\text{Cl}_Y(U\cap Y)=\text{Cl}_X(U\cap Y)\cap Y=\text{Cl}_X(U)\cap Y.$$
Therefore, $$\text{Cl}_X(C)=\text{Cl}_X(\text{Cl}_X(U)\cap Y)\supseteq
\text{Cl}_X(U\cap Y)=\text{Cl}_X(U)=\text{Cl}_X(\text{Cl}_X(U))$$
$$\supseteq\text{Cl}_X(\text{Cl}_X(U)\cap Y)=\text{Cl}_X(C),$$
so $\text{Cl}_X(C)=\text{Cl}_X(U)$ which is regular closed. $\square$
Claim: If $C\in R(X)$, then $C\cap Y\in R(Y)$
Proof: Since $C\in R(X)$, we have $C=\text{Cl}_X(U)$ for some open $U\subseteq X$. Therefore, $$R(Y)\ni\text{Cl}_Y(U\cap Y)=\text{Cl}_X(U\cap Y)\cap Y=\text{Cl}_X(U)\cap Y=C\cap Y.$$ $\square.$
If $X$ is a topological space and $Y\subseteq X$ is dense, then let
$i_{Y,X}:R(Y)\rightarrow R(X)$ be the mapping defined by
$i_{Y,X}(C)=\text{Cl}_X(C)$ and define a mapping $j_{X,Y}:R(X)\rightarrow R(Y)$ by letting $j_{X,Y}(C)=C\cap Y$.
Claim: The mapping $i_{Y,X}$ is injective. More generally, if $C,D$ are distinct closed subsets of $Y$, then $\text{Cl}_X(C)\neq\text{Cl}_X(D)$.
Proof: We can assume that
$x_0\in C\setminus D$. Then since $Y\setminus D$ is open in $Y$, there is an open set $U\subseteq X$ where $U\cap Y=Y\setminus D$. Therefore, since
$U\cap D=\emptyset$, we have $U\cap\text{Cl}_X(D)=\emptyset$ as well, so
$x_0\in\text{Cl}_X(C)$, but $x_0\not\in\text{Cl}_X(D)$. Therefore, the mapping $R(Y)\rightarrow R(X),C\mapsto\text{Cl}_X(C)$ is injective. Mehmet Onat observed that injectivity also follows from the fact that $C=\text{Cl}_Y(C)=Y\cap\text{Cl}_X(C)$. $\square$
Claim: If $C\in R(X)$, then $i_{Y,X}(j_{X,Y}(C))=C$. Therefore, the mappings $i_{Y,X},j_{X,Y}$ are inverses.
Proof: Clearly, $\text{Cl}_X(C\cap Y)\subseteq C$. For the converse direction, suppose that $x_0\in C$. Suppose now that $U$ is an open subset of $X$ with $x_0\in U$. Then since $C=\overline{C^\circ}$, we know that
$U\cap C^\circ\neq\emptyset$. Since $U\cap C^\circ\neq\emptyset$ and $U\cap C^\circ$ is open, we know that $U\cap C^\circ\cap Y\neq\emptyset$ since $Y$ is dense in $X$. Therefore, since $U$ is an arbitrary neighborhood of $x_0$, we have $x_0\in\overline{C^\circ\cap Y}\subseteq\overline{C\cap Y}$. Therefore,
$C=\text{Cl}_X(C\cap Y)$. Q.E.D.
Forcing and point-free topology
Let $X$ be a regular space, and let $D$ be the intersection of all open dense subsets of $X$. If $X$ has no isolated points and is $T_1$, then $D$ is empty, but $D$ has virtual points, and $R(X)$ is isomorphic to the lattice of closed subsets of the space $D$. The virtual points of the space $D$ live inside forcing extensions $V[G].$ For example, the Boolean valued model $V^{R[X]}$ always adds points to the space $D$.
If $L$ is a frame, then let $B_L=\{x^{**}\mid x\in L\}=\{x^*\mid x\in L\}$ where $^*$ is the pseudocomplement operation. Then $B_L$ is a complete Boolean algebra which is the point-free analogue to the lattice of all regular open (and also the lattice of regular closed) subsets of $X$. We say that a sublocale $S$ of a frame $L$ is dense if $0\in S$. The frame $B_L$ is also a sublocale of $L$, and $B_L$ is the smallest dense sublocale of $L.$ A frame is fit if and only if each sublocale is the intersection of open sublocales, so in a fit frame $L$, the Boolean algebra $B_L$ is the intersection of all open sublocales of $L$. If $S$ is a sublocale of a frame $L$, then the Heyting operation on $S$ is the same as the Heyting operation on $L$. In particular, if $S$ is a dense sublocale of $L$, then the pseudocomplement operation on $S$ coincides with the pseudocomplement operation on $S$. Therefore, if $S$ is a dense sublocale of a frame $L$, then $B_S=B_L$.
Best Answer
The family $\mathcal{F}=\{ F\in \mathcal{R}( \beta X) :p\in \operatorname{int}_{\beta X}F\}$ is indeed a filterbase; it is a base for the neighbourhood filter at $p$. As noted in the comments there need not be a unique ultrafilter that extends it; for example in $\beta\mathbb{R}$. Take a point $p$ in $\mathbb{R}$ then $\mathcal{F}$ is generated by $\{[p-2^{-n},p+2^{-n}]:n\in\mathbb{N}\}$; you can add $(-\infty,p]$ or $[p,\infty)$ to $\mathcal{F}$ and still have a filter base in $\mathcal{R}$. So we have at least two $\mathcal{R}$-ultrafilters that extend $\mathcal{F}$.
Srivastava's definition is identical to the one in Gillman and Jerison and hence correct: note that if $Z\in\mathcal{A}^p$ and $E\in f^\#\mathcal{A}^p$ then $Z\cap f^{-1}[E]\neq\emptyset$, hence $f[Z]\cap E\neq\emptyset$ as well. The latter implies that $\operatorname{cl}_{\beta Y}f[Z]\cap f^\#\mathcal{A}^p\neq \emptyset$ too. So $\beta f(p)\in\bigcap\{\operatorname{cl}_{\beta Y}f[Z]:z\in\mathcal{A}^p\}$. Next if $q\neq \beta f(p)$ then take $h:\beta Y\to[0,1]$ with $h(q)=1$ and $h(\beta f(p))=0$. Then $h\circ f$ is continuous and $Z=\{x:h(f(x))\le\frac12\}$ is a member of $\mathcal{A}^p$, but $q$ is not in the closure of $f[Z]$, hence $\bigcap\{\operatorname{cl}_{\beta Y}f[Z]:z\in\mathcal{A}^p\}=\{\beta f(p)\}$.
As an answer to the last question: there is already a definition, independent of the uniqueness of the $\mathcal{R}$-ultrafilters.