Understanding the relations between O(g(n)), Θ(g(n)) and Ω(g(n))
Aren't $\Omega(g(n))$ supposed to tell best running time? 


Not at all. It just means that for all large enough $n$, there exist inputs of size $n$ where the algorithm takes time at least proportional to $g(n)$. 


  Can you please explain why $\Omega(n^2)$ is possible for insertion sort when it should be $\Omega(n)$?


The worst-case running time of insertion sort is $\Omega(n^2)$ because for all $n$, there exist inputs of size $n$ where it takes time proportional to $n^2$. 

The running time of insertion sort in general is not always $\Omega(n^2)$, as you observed; it depends on the input. But the worst-case running time (which does not depend on the input, only on $n$) is $\Omega(n^2)$.