One can see this effect qualitatively from Newtonian first principles (as opposed to Hamiltonian or Lagrangian principles) by looking at a degenerate case, when one moment of inertia is very small and the other two are very close to each other.
More specifically, consider a thin rigid unit disk, initially oriented in the xy plane and centred at the origin (0,0,0). We make the "spherical cow" hypotheses that this disk has infinitesimal thickness and mass, but infinite rigidity. On this disk, we place heavy point masses of equal mass M at the points (1,0,0) and (-1,0,0), and light point masses of equal mass m at the points (0,1,0) and (0,-1,0). Here $0 < m \ll M$, i.e. m should be viewed as negligible with respect to M. (The moments of inertia are then $2m, 2M, 2(m+M)$, though we will not explicitly use these moments in the analysis below.)
We now set up the unstable equilibrium by rotating the disk around the y axis. Thus, the light m-masses stay fixed, while the heavy M-masses rotate in the xz plane. This is in equilibrium: there are no net forces on the m-masses, while the rigid disk exerts a centripetal force on the M-masses that keeps them moving in a circular motion on the xz plane.
We can view this equilibrium in rotating coordinates, matching the motion of the M-masses. (Imagine a camera viewing the disk, rotating around the y-axis at exactly the same rate as the disk is rotating.) In this rotating frame, the disk is now stationary (so the m-masses are stuck at $(0,\pm 1,0)$ and the M-masses are stuck at $(\pm 1,0,0)$), but there is a centrifugal force exerted on all bodies proportional to the distance to the y-axis. The m-masses are on the y-axis and thus experience no centrifugal force; but the M-masses are away from the y-axis and thus experience a centrifugal force, which is then balanced out by the centripetal forces of the rigid disk.
Now let us perturb the disk a bit, so that the m-masses and M-masses are knocked a little bit out of position (but keeping the centre of mass fixed at (0,0,0)). In particular, the m-masses are knocked away from the y-axis and now experience a little bit of centrifugal force. On the other hand, the rigid disk forces the light m-masses to remain orthogonal to the heavy M-masses, by exerting tension forces between the masses. In the regime where m is negligible compared to M, these tension forces will barely budge the heavy M masses (which therefore remain essentially fixed at $(\pm 1,0,0)$ in the rotating frame), so the effect of these tension forces is to constrain the m-masses to lie in the yz plane (up to negligible errors which we now ignore). Rigidity also keeps the m-masses at a unit distance from the origin, and antipodal to each other, so the m-masses are now constrained to be antipodal points on the unit circle in the yz plane. However, other than this, rigidity imposes no further constraints on the location of the m-masses, which can then move freely as antipodal points in this unit circle.
The effect of centrifugal force in the rotating frame is now clear: if an m-mass (and its antipode) is perturbed to be a little bit off the y-axis in this unit circle with no initial velocity, then centrifugal force will nudge it a little further off the y-axis, slowly at first but with inexorable acceleration. Eventually it will shoot across the unit circle and then approach the antipode of its previous position. At this point the centrifugal forces act to slow the m-masses down, reversing all the previous acceleration, until one ends up with no velocity at a small distance from the antipode. The process then repeats itself (imagine a marble rolling between two equally tall hills, starting from a position very close to the peak of one of the hills).