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.

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.

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 in 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.

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.