You're confusing all forces labelld $\frac{mv^2}{r}$ in the diagram as centripetal forces. A force labelled $\frac{mv^2}{r}$ can be one of three things:
1. Centripetal force
2. Centrifugal force
3. A different balancing force that is calculated to be equal to $\frac{mv^2}{r}$ (This is not exactly different from the other two)
Centrifugal forces
I recently explained the concept of a psuedoforce over here. See the last part of the answer (labelled "psuedoforces") if you do not know what they are before reading the next part.
The centrifugal force is basically the psuedoforce acting in a rotating frame. Basically, a frame undergoing UCM has an acceleration $\frac{mv^2}{r}$ towards the center. Thus, an observer in that rotating frame will feel a psuedoforce $\frac{mv^2}{r}$ outwards. This psuedoforce is known as the centrifugal force.
Unlike the centripetal force, the centrifugal force is not real. Imagine a ball being whirled around. It has a CPF $=\frac{mv^2}{r}$, and this force is the tension in the string. But, if you shift to the balls frame (become tiny and stand on it), it will appear to you that the ball is stationary (as you are standing on it. The rest of the world will appear to rotate). But, you will notice something a bit off: The ball still has a tension force acting on it, so how is is steady? This balancing of forces you attribute to a mysterious "centrifugal force". If you have mass, you feel the CFF, too (from the ground, it is obvious that what you feel as the CFF is due to your inertia)
The diagram
Now, they have solved the problem in the biker's frame. Thus, there is an outwards CFF$=\frac{mv^2}{r}$ acting on the center of mass. Now, the biker stays stationary wrt the rotating frame in the horizontal direction (this is the radial direction in the ground frame). Thus, there is some other force, balancing the CFF. This if the friction force, and it has been labelled confusingly as $\frac{mv^2}{r}$ (at the wheel). Without it, in the rotating frame, the body would not be at rest/uniform motion.
In the ground frame, the friction force is the CPF.
Long story short
Neither of the $\frac{mv^2}{r}$s in your diagram are being considered as CPFs. The one at center of gravity is CFF, and the other one at the wheel is a friction force balancing the CFF in rotating frame. It is also the CPF in the ground frame.
The equation implies that if I push on an object with force $\vec F$, it will start moving at velocity $\vec v = \vec F / R$.
You are correct when you say that having vectors on either side of the equal sign means they will be in the same direction. If I push an object away from me, this equation says it will move in the direction I pushed it. It also says that stronger input forces give faster output velocities. If I just give a small push, the object will move not so quickly, but a big push means fast motion. So far it sounds pretty reasonable, right?
There are several ways this equation doesn't accurately describe motion, though. I will take one illustrative example.
Say an object is sitting in front of me (in space, so there isn't any gravity or friction). I give this object a little push and it starts moving. After my push is done, I'm not imparting any more force to the object. What does this equation say should happen, and what really happens?
This equation says that when my hand is in contact and I am pushing the object, it will move with velocity $\vec v$. But as soon as I take my hand away, the force goes to $0$, therefore the object's velocity goes to $0$ as well. That means the object only moves when I push it and immediately stops dead when I am done pushing.
Of course, we know that in reality if I push an object it will continue moving after the push is done. And in my example, in space with no gravity and no friction, the object will continue moving at a constant velocity forever. That's because in the real world, forces don't cause velocities, forces change velocities; no force, no velocity change.
Best Answer
You were probably expected to note that the path of any point of the tire is a cycloid. At the point of contact with the ground, the tire is not moving. As it rises from the ground it is moving faster, with a speed that is $v(1+\sin \theta)$ where $v$ is the bike speed and $\theta$ is measured counterclockwise from horizontal. The centrifugal force is rising as $\theta$ decreases from $270^\circ$ and at some point it exceeds the strength of the attachment to the wheel. When it lets go it should get the tangential velocity of that point of the wheel, which is not the same as the speed of the bike.