big O of a complex function
I'm assuming you meant $l(x) = a^{b-x}/h(x)$, where $a \geq 1$. This term is $o(1)$ so can be ignored. For the main term, we cannot conclude anything since we only have an upper bound on $h$; we would like a lower bound on $h$. It could be, for example, that $g(k) = \log k$ and $h(k) = e^{-k}$, a function certainly satisfying $h(k) = O(k)$, and in that case $f(x) = \Omega(e^x)$. 

Assuming you meant $h(k) = \Theta(k)$, or at least $h(k) = \Omega(k)$, we can estimate the sum using an integral:
$$ \sum_{k=0}^x \frac{g(k)}{h(k)} = O\left(\sum_{k=0}^x \frac{\log k}{k}\right) = O\left(\int_0^x \frac{\log k}{k} dk\right) = O(\log^2 x),$$
using $\int \frac{\log k}{k} = \frac{1}{2} \log^2 k$. If $g(k) = \Theta(\log k)$ and $h(k) = \Theta(k)$ then $f(x) = \Theta(\log^2 x)$.