In polar coordinates (in the stationary, outside system), which is the best choice of coordinate system for circular motion investigation, we have the newton's second law like this (obtained by differentiating a general motion twice):
$$\mathbf{F}=m(-r\dot\theta^2 \mathbf{\hat{r}}+\ddot r\mathbf{\hat r}+2\dot r\dot \theta \boldsymbol{\hat \theta}+r\ddot \theta\boldsymbol{\hat \theta})$$
So It is no longer like $\mathbf{F}=m(\ddot x \mathbf{\hat x}+\ddot y \mathbf{\hat y}+\ddot z \mathbf{\hat z})$.
Now. If $r$ is not changing, and $\dot\theta$ (angular velocity) is constant, we conclude $\dot r=\ddot \theta=0$. So we can simplify the general case and it becomes this:
$$\mathbf{F}=-mr\dot\theta^2 \mathbf{\hat{r}}$$
You can see that there is only one force (called centripetal force) acting on the moving particle. Which is radially inward.
But what about centrifugal one??
There is no force like centrifugal force. But! if you go to the system of coordinates of the moving particle, which means you are the moving particle. Then, because your system is not performing a uniform linear motion, there must be a fictitious force on your system. Since the only force on your system was the centripetal one we just calculated, the centrifugal one must be minus that force, which now becomes radially outward.
That our choice of system of coordinates which can totally change the math.