Why does the frictional centripetal force create a moment around a leaning bike? 
Image source: https://geekswipe.net/science/physics/why-bike-lean-in-turn/
This diagram, along with several others I have seen, seem to use the centre of gravity as the pivot when calculating the moment/torque acting on the biker.
This is why the centripetal force creates a counter clockwise moment on the biker.
However, aren't the tyres of the bike (which are in contact with the ground) the pivot(s), considering they are what the motion is centred around? In that case, doesn't the centripetal force not create a moment because the distance between it and the pivot is 0?   
 A: Moment can be calculated with respect to any point you choose, since you may fix any point you want to check how things rotate about it. You just need to be careful with your interpretation of what that calculation means.
Indeed, with respect to the point of contact with the ground, the tyres don't create a moment, but they are not the whole story.
As an analogy, think of a bar connected to an axis such that it can be freely rotated. When you push the bar to rotate it, you create a torque with respect to the axis of rotation. But when you switch your reference point to the point of contact of your hands with the bar, you will feel the force and torque that the now rotating axis (it rotates around you in your rest frame) applies on your hands, and these will be the relevant quantities to use in your diagrams.
Going back to the bike, when considering a rotating reference frame, you need to take into account the centrifugal forces that emerge in that frame, which in this case arise due to fact that the upper part of the biker+bike system "tries" to move in a straight line with nothing but gravity to balance it.
