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.
The conjectured formula can be proved by induction on $\mathrm{wt}(n)$.
For $\mathrm{wt}(n)=0$, we have $n=0$ and the conjectured formula trivially holds.
Now, for a positive integer $\ell$, suppose that the conjectured formula holds for all $\mathrm{wt}(n)<\ell$. Let us prove it for $\mathrm{wt}(n)=\ell$. So, let $n=2^{t_1}(1+2^{t_2+1}(1+\dots(1+2^{t_\ell+1}))\dots)$ with $t_j\geq 0$ and $n'=2^{t_2}(1+\dots(1+2^{t_\ell+1}))\dots)$, so that
\begin{split}
a(n) &= a(2^{t_1}(2n'+1)) = \sum_{m=0}^{t_1} \binom{t_1+1}{m} a(2^kn') \\
&=\sum_{m=0}^{t_1} \binom{t_1+1}{m} \sum_{j=0}^{2^{\ell-1}-1} (-1)^{\ell-1-\mathrm{wt}(j)} (1+\mathrm{wt}(j))^{m+t_{2}+1} \prod_{k=2}^{\ell} (1+\mathrm{wt}(\lfloor \tfrac{j}{2^{k-1}}\rfloor))^{t_{k+1}+1} \\
&=\sum_{j=0}^{2^{\ell-1}-1} (-1)^{\ell-1-\mathrm{wt}(j)} \left[(2+\mathrm{wt}(j))^{t_{1}+1} - (1+\mathrm{wt}(j))^{t_{1}+1}\right] \prod_{k=1}^{\ell} (1+\mathrm{wt}(\lfloor \tfrac{j}{2^{k-1}}\rfloor))^{t_{k+1}+1} \\
&=\sum_{j=0}^{2^{\ell-1}-1} \left[(-1)^{\ell-\mathrm{wt}(2j+1)}\prod_{k=0}^{\ell} (1+\mathrm{wt}(\lfloor \tfrac{2j+1}{2^k}\rfloor))^{t_{k+1}+1} + (-1)^{\ell-\mathrm{wt}(2j)} \prod_{k=0}^{\ell} (1+\mathrm{wt}(\lfloor \tfrac{2j}{2^k}\rfloor))^{t_{k+1}+1}\right] \\
&=\sum_{j'=0}^{2^\ell-1} (-1)^{\ell-\mathrm{wt}(j')} \prod_{k=0}^{\ell} (1+\mathrm{wt}(\lfloor \tfrac{j'}{2^k}\rfloor))^{t_{k+1}+1}, \\
\end{split}
where $j'$ corresponds to values $2j$ and $2j+1$ (in other words, $j=\lfloor j'/2\rfloor$). QED
Best Answer
The same idea of grouping terms by the number of unit bits, as well as grouping by the value of $\mathrm{wt}(j)$ (representing $j$ via individual bits) works here: \begin{split} s(n,m) &= \sum_{\ell=0}^n \sum_{t_1 + \dots + t_\ell \leq n-\ell} \sum_{j_1,\dots,j_\ell\in\{0,1\}} m^{\ell-\sum_{i=1}^\ell j_i} \prod_{k=1}^{\ell} (1+\sum_{i=1}^k j_i)^{t_k+1} \\ &=[x^{n+1}]\ \sum_{\ell=0}^n \sum_{j_1,\dots,j_\ell\in\{0,1\}} m^{\ell-\sum_{i=1}^\ell j_i} \prod_{k=0}^{\ell} \frac{(1+\sum_{i=1}^k j_i)x}{1-(1+\sum_{i=1}^k j_i)x} \\ &=[x^{n+1}]\ \sum_{\ell=0}^n \sum_{J=0}^\ell \sum_{j_1,\dots,j_\ell\in\{0,1\}\atop j_1+\dots+j_\ell=J} m^{\ell-J} \prod_{k=0}^{\ell} \frac{(1+\sum_{i=1}^k j_i)x}{1-(1+\sum_{i=1}^k j_i)x} \\ &=[x^{n+1}]\ \sum_{\ell=0}^n \sum_{J=0}^\ell m^{\ell-J} \sum_{d_0,d_1,\dots,d_J\geq 1\atop d_0+d_1+\dots+d_J=\ell+1}\prod_{k=1}^{J+1} \left(\frac{kx}{1-kx}\right)^{d_k} \\ &=[x^{n+1}]\ \sum_{\ell=0}^n m^{\ell+1}[y^{\ell+1}]\ \sum_{J=0}^\ell \prod_{k=1}^{J+1} \frac{kxy}{(1-kx-kxy)m} \\ &=[x^{n+1}]\ \sum_{J=0}^n \prod_{k=1}^{J+1} \frac{kx}{1-kx-kmx} \\ &=\sum_{J=0}^n (J+1)!\ [x^{n-J}]\ \prod_{k=1}^{J+1} \frac{1}{1-k(m+1)x} \\ &=\sum_{J=0}^n (J+1)!\ (m+1)^{n-J} \left\{n+1\atop J+1\right\}. \end{split} It also shows that the summands in the last formula corresponds to fixed values of $\mathrm{wt}(j)=J$.