The Earth-Moon system is best thought of as the Moon orbiting the Earth, but not around the Earth's centre, rather their barycentre (the common centre of mass). This means that the Moon orbits the Earth but pulls the Earth around a little so that the Earth has a translational motion around the barycentre as motion in a circle (but not a rotation/revolution about that point). If the Earth and Moon both revolved around a common centre then they would be mutually tidally locked and both would show each other their same face forever, with each not changing position in the sky of the other at all (as is the case for Pluto and Charon, which are a binary dwarf planetary system). This can be appreciated in animations contained in an article by Paolo Sirtoli at https://www.vialattea.net/content/tides-and-centrifugal-force/.
A centrifugal force is a fictitious force that is introduced into a non-inertial reference frame to make sense of observations for someone inside that non-inertial reference frame. So by definition, there is no centrifugal force that can be measured.
If the Earth's centre is to be considered as a reference frame then it is non-inertial because it is being pulled on by the moon, and so to make it quasi-inertial, a fictitious inertial force (since the word centrifugal implies rotation, which is not happening here) is introduced, which is everywhere, i.e. uniform, a force that opposes the Moon's pull as at the centre of the Earth. The superposition of the Moon's varying gravity force across the diameter of the Earth and this fictitious inertial force produces the classic tidal force diagram that you see, that looks to produce the two tidal bulges, on the near and far side of the Moon. In reality the tidal force is so small in comparison to the Earth's gravity force (by some ten million times) that it is not this force that causes the rise in the tides, rather the forces at ~45 degree offsets that serve to slide the water surface perpendicularly to the gravity force - here the small tidal force has no opposing gravity force - so as to collectively push up a noticeable tide at the near and far sides.
Alternatively, the reference frame can be thought of as the barycentre, which is truly fixed in relation to the inertial stars (ignoring other influences). In this case, the reference frame is truly inertial, but the Earth's centre actually translates about the barycentre and so a real centripetal force has to be taken into account, pointing towards the barycentre. Now, if you analyse the motion of the translating Earth (centre) about the barycentre carefully, you will notice that all points on the Earth are moving in a circular motion about their own (bary)centres, with a single radius value, and that these forces all point towards their own (bary)centres in a direction towards the distant Moon, i.e. it is a uniform centripetal force that acts in the region of the Earth, but towards the Moon. This too is evident in Paolo Sirtoli's article. This centripetal force is provided by the Moon's varying gravitational force across the Earth's diameter and when the two are compared, the result is the same (net) tidal forces acting in the same direction to produce the near and far side bulges (via the forces acting at the ~45 degree offsets, as before).
Now, you can consider a frame of reference in which things rotate around the barycentre, and so a centripetal force arises again, which might throw up a difference between the near side and far side centripetal force, which you might think might affect the tides. However, you need to 'convert' your barycentre reference origin to the centre-of-earth reference origin, and when you do this you find that the centripetal forces are transformed into uniform forces around the centre of the Earth, and so no differential arises that could affect tides.
You may note that the Earth's gravity and axial rotation play no part in tides, - these forces are from within the Earth and so cannot exert a force on the Earth.