Let $n=m^tk$ where $m\nmid k$. Then $f(n)=m^t$.
Furthermore, if $t>0$, then $f(n/m)=m^{t-1}$ and $n-f(n/m)=m^{t-1}(mk-1)$. It follows that $a(n)=a(m^{t-1}k)+a(m^{t-1}(mk-1))$ and further by induction on $t$,
$$
(\star)\qquad a(n)=\sum_{i=0}^t \binom{t}{i} a\big(m^ik-\frac{m^i-1}{m-1}\big).
$$

**CASE $m>2$.** In this case, formula $(\star)$ reduces to
$$a(n) = a(k)+(2^t-1)a(k-1).$$

Let's analyze $s(n)$.
It is clear that for any $\ell\not\equiv 0\pmod{m}$, we have
$$\sum_{k=0\atop k\equiv \ell\pmod{m}}^{m^n-1} a(k) = s(n-1)$$
and correspondingly
$$\sum_{k=0\atop k\equiv 0\pmod{m}}^{m^n-1} a(k) = s(n) - (m-1)s(n-1)$$

Now we are ready to derive a recurrence for $s(n)$ by grouping the summation indices based on the power $m^t$ they contain:
\begin{split}
s(n) &= 1 + \sum_{t=0}^{n-1} \sum_{k=1\atop k\not\equiv 0\pmod{m}}^{m^{n-t}-1} \left(a(k) + (2^t-1)a(k-1)\right) \\
&=1 +\sum_{t=0}^{n-1} \left((m-1)s(n-t-1) + (2^t-1)((m-2)s(n-t-1) + s(n-t)-(m-1)s(n-t-1))\right) \\
&=1 +\sum_{t=0}^{n-1} \left((m-2^t)s(n-t-1) + (2^t-1)s(n-t)\right) \\
&=2 - 2^n + \sum_{t=1}^n (2^{t-1}+m-1)s(n-t).
\end{split}

Restating the above recurrence in terms of the generating function $S(x):=\sum_{n\geq 0} s(n)x^n$, we have
$$S(x) = \frac{2}{1-x} - \frac{1}{1-2x} + \left(\frac{x}{1-2x} + \frac{(m-1)x}{1-x}\right)S(x).$$
That is,
$$S(x) = \frac{1-3x}{1 - (m+3)x + (2m+1)x^2},$$
from where the required recurrence follows instantly.

**CASE $m=2$.** In this case, formula $(\star)$ takes form:
$$a(n)=a(k)+\sum_{i=1}^t \binom{t}{i} a(2^{i-1}(k-1)).$$
It further follows that for $n=2^{t_1}(1+2^{1+t_2}(1+\dots(1+2^{1+t_\ell}))\dots)$ with $t_j\geq 0$, we have
\begin{split}
a(n) &= \sum_{i_1=0}^{t_1} \binom{t_1}{i_1} \sum_{i_2=0}^{t_2+i_1} \binom{t_2+i_1}{i_2} \sum_{i_3=0}^{t_3+i_2} \dots \sum_{i_\ell=0}^{t_\ell+i_{\ell-1}} \binom{t_\ell+i_{\ell-1}}{i_\ell} \\
&=\prod_{j=1}^\ell (\ell+2-j)^{t_j}.
\end{split}

Grouping the summands in $s(n)$ by the number of unit bits, we have
\begin{split}
s(n) &= \sum_{\ell=0}^n\ \sum_{t_1+t_2+\dots+t_{\ell}\leq n-\ell}\ \prod_{j=1}^\ell (\ell+2-j)^{t_j}\\
&= \sum_{\ell=0}^n\ \sum_{t_1+t_2+\dots+t_{\ell}+t_{\ell+1} = n-\ell}\ \prod_{j=1}^{\ell+1} (\ell+2-j)^{t_j}\\
&=\sum_{\ell=0}^n [x^{n-\ell}]\ \prod_{j=1}^{\ell+1} \frac1{1-jx} \\
&=\sum_{\ell=0}^n S(n+1,\ell+1) \\
&=B_{n+1},
\end{split}
where $S(n+1,\ell+1)$ are Stirling numbers of second kind, and $B_{n+1}$ is Bell number.

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