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 $t$,
\begin{split}
a(n)&=\sum_{i=0}^t \binom{t}{i} a\big(m^ik-\frac{m^i-1}{m-1}\big)\\
&=a(k)+(2^t-1)a(k-1).
\end{split}

In particular, if $k\not\equiv 1\pmod{m}$, we have $a(n)=2^ta(k)$.

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Now, 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.