In the process of gyroscopic precession:
For the gyro wheel there is continuous internal relocation of (angular) momentum. That is to say, there are continuous bending forces as some areas of the gyro wheel are being accelerated and others are being decelerated.
There is a case that also has internal relocation of (angular) momentum, but easier to understand. It's a case that is commonly referred to as 'Feynman's wobbling plate'.
When a spinning plate is thrown up in the air the angular momentum of the plate will be constant; constant both in magnitude and direction.
Any wobble of that spinning plate will not go away when the plate is free from the throwing hand; if that plate would be in a vacuum (hence no air friction) then the wobbling will sustain forever. Youtube video by Jacob Forster: Feynman's wobbling plate
The fact that the plate is rigid is an essential element. It is because of that rigidity that the bending forces do not give rise to dissipation of energy. Comparison: making pizza dough stretch by spinning it (giving the dough freedom to stretch by throwing it upward). You never see that spinning pizza dough wobble. That's because the pizza dough is not rigid; any bending force immediately dissipates.
In the case of a gyroscope:
A necessary condition for gyroscopic precession to occur is that the gyro wheel is rigid. It has to be rigid because in order to display gyroscopic precession the gyro wheel has to support internal relocation of momentum.
There is a 2012 answer by me in which I describe the mechanism of gyroscopic precession. That description explains why it is that when a spinning gyroscope is released it doesn't flip down, but instead settles to gyroscopic precession.
In that description I do not use the abstract concept of spin angular momentum. Instead the explanation capitalizes on symmetry.