Suppose $x\in\mathbb{Q}$, $0\lt x\lt1$, and $x$ has the base-$p$ expansion
$$
x=\sum_{k=1}^\infty\frac{d_k}{p^k}\tag{1}
$$
Then
$$
\frac{\{p^nx\}}{p^n}=\sum_{k=n+1}^\infty\frac{d_k}{p^k}\tag{2}
$$
So that
$$
\begin{align}
f_p(x)
&=\sum_{n=0}^\infty\frac{\{p^nx\}}{p^n}\\
&=\sum_{n=0}^\infty\sum_{k=n+1}^\infty\frac{d_k}{p^k}\\
&=\sum_{k=1}^\infty\sum_{n=0}^{k-1}\frac{d_k}{p^k}\\
&=\sum_{k=1}^\infty\frac{k\,d_k}{p^k}\tag{3}
\end{align}
$$
Since the sum in $(3)$ starts at $k=0$, $f_p(x)-x$ is the function in the question. However, if $f_p(x):\mathbb{Q}\mapsto\mathbb{Q}$, then $f_p(x)-x:\mathbb{Q}\mapsto\mathbb{Q}$.
Finite base-$p$ expansion
Obviously, if the base-$p$ expansion of $x$ is finite, then the sum in $(3)$ is finite
$$
f_p(x)=\sum_{k=1}^m\frac{k\,d_k}{p^k}\tag{4}
$$
which is a finite sum of rational numbers, hence $f_p(x)\in\mathbb{Q}$.
Repeating base-$p$ expansion
If the base-$p$ expansion of $x$ repeats with period $m$, then
$$
\begin{align}
f_p(x)
&=\sum_{k=1}^m\sum_{n=0}^\infty\frac{(k+nm)d_k}{p^{k+nm}}\\[6pt]
&=\sum_{k=1}^m\frac{d_k}{p^k}\sum_{n=0}^\infty\frac{k+nm}{p^{nm}}\\[6pt]
&=\sum_{k=1}^m\frac{d_k}{p^k}\left(\frac{kp^m}{p^m-1}+\frac{mp^m}{(p^m-1)^2}\right)\\[6pt]
&=\frac1{p^m-1}\left(mx+p^m\sum_{k=1}^m\frac{k\,d_k}{p^k}\right)\tag{5}
\end{align}
$$
which is again a finite sum of rational numbers, hence $f_p(x)\in\mathbb{Q}$.
Mixed base-$p$ expansions
Note that if there are no base-$p$ carries when adding $x$ and $y$, then each digit of the sum is the sum of the digits, and therefore, by $(1)$ and $(3)$,
$$
f_p(x+y)=f_p(x)+f_p(y)\tag{6}
$$
Furthermore,
$$
\begin{align}
f_p\left(\frac{x}{p^n}\right)
&=\sum_{k=1}^\infty\frac{(k+n)d_k}{p^{k+n}}\\
&=\frac1{p^n}\left(nx+f_p(x)\right)\tag{7}
\end{align}
$$
Combining $(4)$, $(5)$, $(6)$, and $(7)$, we get
Conclusion
If $x\in\mathbb{Q}$, then
$$
\sum_{k=1}^\infty\frac{\{p^kx\}}{p^k}=f_p(x)-x\in\mathbb{Q}
$$
Example 1
In base $5$, $\frac{14}{25}=.\color{#C00000}{24}$. By $(4)$
$$
\begin{align}
f_5\left(\frac{14}{25}\right)
&=\frac{\color{#00A000}{1}\cdot\color{#C00000}{2}}{5^{\color{#00A000}{1}}}+\frac{\color{#00A000}{2}\cdot\color{#C00000}{4}}{5^{\color{#00A000}{2}}}\\[6pt]
&=\frac{18}{25}
\end{align}
$$
Example 2
In base $5$, $\color{#0000FF}{\frac13}=.\overline{\color{#C00000}{13}}$, therefore, $p=5,m=2,d_1=1,d_2=3$. By $(5)$
$$
\begin{align}
f_5\left(\color{#0000FF}{\frac13}\right)
&=\frac1{5^2-1}\left(2\cdot\color{#0000FF}{\frac13}+5^2\left(\frac{\color{#00A000}{1}\cdot\color{#C00000}{1}}{5^\color{#00A000}{1}}+\frac{\color{#00A000}{2}\cdot\color{#C00000}{3}}{5^\color{#00A000}{2}}\right)\right)\\[6pt]
&=\frac{35}{72}
\end{align}
$$
Example 3
In base $5$, $\frac{67}{75}=.24\overline{13}$. Using $(6)$ and $(7)$ and the previous examples, we get
$$
\begin{align}
f_5\left(\frac{67}{75}\right)
&=f_5\left(\frac{14}{25}\right)+f_5\left(\frac13\cdot\frac1{25}\right)\\[6pt]
&=\frac{18}{25}+\frac1{5^2}\left(2\cdot\frac13+\frac{35}{72}\right)\\[6pt]
&=\frac{1379}{1800}
\end{align}
$$