$\newcommand\ep\varepsilon$The answer is still no (assuming you wanted $c>0$). Indeed, let $$s_n:=\sum_{k=0}^n a_k\sin(k\ep),\quad t_n:=\sum_{k=0}^n b_k\sin(k\ep),$$ where $\ep=3\pi/4$, $a_0=b_0=10$, $a_1=1$, $b_1=9$, and $a_k=b_k=2^{-k}$ for $k\ge2$.

Then $c_1a_k\le b_k\le c_2a_k$ for $c_1=1$, $c_2=9$, and all $k\ge0$. Moreover, $f(n):=s_n>0$ for all $n\ge0$.

However, $t_1=-9 + 5\sqrt2<-19/20<0$ and, moreover, $t_n<0$ for all $n\ge2$.

$\newcommand\ep\varepsilon$The answer is still no (assuming you wanted $c>0$). Indeed, let $$s_n:=\sum_{k=0}^n a_k\sin(k\ep),\quad t_n:=\sum_{k=0}^n b_k\sin(k\ep),$$ where $\ep=3\pi/4$, $a_0=b_0=10$, $a_1=1$, $b_1=9$, and $a_k=b_k=2^{-k}$ for $k\ge2$.

Then $c_1a_k\le b_k\le c_2a_k$ for $c_1=1$, $c_2=9$, and all $k\ge0$. Moreover, $s_0>0$ and $s_1=-1 + 5\sqrt2>6>0$, whence $s_n>6-\sum_{k=2}^\infty 2^{-k}>0$ for $n\ge2$. So, $s_n>0$ for all $n\ge0$. So, letting $f(n):=s_n$ for all $n\ge0$, we get a positive function $f$ such that $s_n\le f(n)$ for all $n\ge0$. Thus, all your conditions on the $a_k$'s and $b_k$'s are satisfied.

However, $t_1=-9 + 5\sqrt2<-19/20<0$ and, moreover, $t_n<-19/20+\sum_{k=2}^\infty 2^{-k}<0$ for all $n\ge2$. Thus, the condition $t_n\le cf(n)$ fails to hold for any $c>0$ and any $n\ge1$.