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} $