I) Here we will give a qualitative rather than quantitative analysis. Consider first the axially symmetric 3D tippe top, which means that there will be two principal axis with maximal moment of inertia. (If we wish, we can for simplicity model the top as a spherically symmetric ball, with a heavy point mass off the geometric center $O$.) Empirically, we observe to a first approximation that:
The angular vector $\vec{\omega}$ is mostly vertical and completes a small precession for each revolution of the top. (Let's assume that $\vec{\omega}$ is upwards rather than downwards.)
The rotation of the top is around a point $Q$ somewhere between the geometric center $O$ and the center of mass $C$.
The point $P$ of contact completes a small circular orbit for each revolution of the top.
The top is mostly rolling without sliding. In other words, to a first approximation, it is static friction rather than kinetic friction, and hence the mechanical energy is conserved.
The angular momentum vector $\vec{L}$ will mostly point upwards but tilted a bit to the same (opposite) side as $C$ iff $O$ is above (below) $C$. In other words, the angular momentum vector $\vec{L}$ completes a small precession for each revolution of the top. We will assume that this precession of $\vec{L}$ (and the corresponding net torque needed for this precession) remains very small during the entire inversion process.
Now the torques of gravity and the normal force work in unison to increase the precession of $\vec{L}$. The cancellation of this torque can only come from the friction force acting in the same direction (horizontally speaking) as where the center of mass $C$ is.$^1$
Naturally, due to an imperfect rolling motion, the friction force will then tend to slide the point of contact in the direction of the friction force, thereby raising the center of mass $C$. (Del Campo showed that sliding friction must play an essential role in the inversion process, cf. Ref. 1. Of course sliding friction is also responsible for that the top eventually comes to rest and looses its mechanical energy.)
II) The asymmetric 2D discs$^2$ works similarly, although now the rotation will mainly be around the unstable principal axis with intermediate moment of inertia, leading to the tennis racket/Dzhanibekov effect, as correctly pointed out by ja72 in his answer, see also e.g. this Phys.SE post and links therein. In situations without friction, the tennis racket/Dzhanibekov effect explains the last unresolved part of the solution video.
References:
- Richard J. Cohen, The tippe top revisited, http://dx.doi.org/10.1119/1.10926 (hat tip: Leonida)
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$^1$ Note that the friction torque argument in this Basic Saucer Physics 101 video seems to lead to the opposite effect where the center of mass $C$ gets lowered (rather than raised).
$^2$ The geometric details of the asymmetric 2D discs is to a large extent irrelevant for the effect.