There's nothing wrong with JEB's answer, but I think this may help:
You write
Now I know that centrifugal and centripetal force are opposite to each other
...
But then why... the rotating object faces a centrifugal force outwards?
Centrifugal and centripetal forces always point in opposite directions, along the radius away from or towards the center of rotation. However, they need not be equal in magnitude. The expression you found for centripetal force is only identical to the expression for centrifugal force (except for the direction) for the special case of an object for which there is no acceleration in the radial direction in the frame which measures the centrifugal force. In other cases, the $r$ means something different in the two expressions. See the note at the end if interested.
To explain in more detail:
Let us take the frame of an observer who is rotating with angular velocity $\omega$ about the point $r=0$. For instance, we can attach the observer to a big rotating wheel.
If we attach a relaxed spring to a weight on one end and to the spinning wheel on the other end, with initial velocity 0 in the radial direction, as measured from a frame corotating with the wheel, we will start out by measuring a centrifugal force on the weight and zero centripetal force. The weight accelerates in the radial direction. This gives it a velocity in the radial direction, which makes it increase in radius. As it increases in radius, the inward pointing (centripetal) force of the stretched spring increases.
Eventually, the system will reach an equilibrium in which there is no longer any acceleration in the radial direction. This will be the amount of stretch at which the centripetal force equals the centrifugal force in the co-rotating frame.
If the body is accelerating in the co-rotating frame, the $r$ in the canonical expression for centripetal force does not indicate the distance to the center of rotation of the co-rotating frame.