Why is $\Theta$ notation suitable to insertion sort to describe its worst case running time?
You are confusing two different notions.
The definition of $\Theta$ that you give is correct, and indeed the running time of insertion sort, in the worst case, is $\Theta(n^2)$, since it has a quadratic running time.

The fact that this is the worst running time is somewhat irrelevant here. You have some function, and you are classifying it as $\Theta(n^2)$, because you can.

Indeed, the runtime of the algorithm in the best case can be a different function, which may or may not be $\Theta(n^2)$.

It may make more intuitive sense to just say that in the worst case, the runtime is $O(n^2)$, which implies that in all the cases the runtime is $O(n^2)$, but observe that this is also implied by the stronger $\Theta$ notations.

As an illustrative example, consider this: You can get home in two paths, one takes 10 km, and the other 3 km, but the 3 km is only open on certain hours of the day.

You can say the following: when you walk home, in the worst case you will walk exactly 10 km. This is true, and it implies that you always walk at most 10 km. But the former is a stronger claim, so why not use it?