After reading on many different opinions and answers I have obtained the right intuition. I am sharing it here just so that it may help anyone who stumbles on it again.
Disclaimer : I am gonna build this answer directly over flevin's thought experiment. Please read it if you haven't already.
I am gonna completely ignore the radial component as the intuition for it is fairly obvious. In the discussion that follows I will exclude it.
Now, the tangential component of acceleration is best seen as composed of not 2 but three different terms (stay with me for a sec, I will clarify) -
$$a_t = (\alpha r) + (\omega \dot{r}) + (\omega \dot{r})$$
Of these 3 terms, the first term is fairly obvious and intuitive. So let's drop it by saying it that the body carries no angular acceleration (taking hint from @flevinBombastus' thought experiment). Now the expression becomes -
$$a_t (\text{at constant} \space \omega) = 0 + (\omega \dot{r}) + (\omega \dot{r})$$
Now, as @flevinBombastus' rightly argued in his thought experiment, as the radial distance of particle changes from origin, it's tangential velocity must also change. Intuitively (hell, also rigorously) for $\Delta r$ change in radius, tangential velocity changes by $\omega \Delta r$. Thus, we need an tangential acceleration of $\omega \dot{r}$ to bring about this change. This explains the second term in our expression. Thanks to @flevinBombastus' for the thought experiment which gave me this wonderful idea.
But wait, it seems like we accounted for everything already, so from where does that last $\omega \dot{r}$ pop up from? That's the tricky part, but absolutely not non-intuitive. Here's the big idea -
Let's ask ourselves, what's the difference between a purely uniform circular motion versus a motion in which we slowly reel out string as described in the flevin's thought experiment? Answer : It's the radial velocity. It is absent in case of circular motion but obviously present in the present case under study. So, our particle has this radial velocity and if you think about it, the radial velocity is a rotating vector. But, and here is the essence of argument, if radial velocity is a rotating vector it implies that we need another tangential acceleration to change it's direction! Now, it can be shown by explicit calculation that the acceleration required to bring about this rotation is equal in magnitude to $\omega \dot{r}$ !!!!
That means, that means ... that the third term of our expression is also unveiled.
TL;DR, there is no factor of 2 in the "coriolis" acceleration. It is actually made up of 2 terms arising from entirely different contexts - one to bring about the change in tangential velocity arising due to radial movement of particle and the other to bring about rotation of radial velocity vector. It just so happens, I like to say, that by coincidence the magnitude of both turned out to be same.