In this question, the answer given discusses that fluorine has a good orbital overlap with hydrogen and also that fluorine is not very polarizable. However, is it always true that strong bonding is inversely related to polarizability?
Fluorine is not very polarizable because it is small. Its electrons, therefore, are all close together. A polarized atom has shoved all its electrons to one side. Since they are close together in Fluorine, the negative-negative repulsion is too large for that. In another halide such as Iodine, whose valence electrons are much farther from each other, much better polarization can be achieved.
More polarizability does NOT always mean better or weak bonding. Generally, polarizable atoms like to bond to each other, and non-polarizable atoms like to bond to other non-polarizable atoms. For more info, see:
The polarisability is a measure of how an atom's electrons can be displaced from equilibrium by a nearby electric field.
The polarisability is approximately inversely proportional to the ionisation energy (taken across periods in the periodic table); the easier it is to ionise the easier it is to alter the average electron's positions because they are less energetically bound to the nucleus.
Forming a chemical bond is an all together different thing and depends on the balance of the electron's kinetic and potential energies and these are determined by electron-electron repulsion, electron-proton attraction and proton-proton repulsion energies. These energies are calculated using a quantum mechanical approach that involves Coulomb and related integrals.
It is not true that the strength of a bond is inversely proportional to the polarisability of the atoms involved in the bond.
Bond strength involves various factors, including the electronegativity difference between the atoms, the orbitals involved in bonding and electrostatic interactions. The list is non-exhaustive and I don't think all the factors can be easily summarised in a list even.
In the post you have linked in your question, "low polarisability" of the fluorine atom was discussed as a reason for the strong hydrogen-fluorine bond. Now, let us make sense of that...
When an atom is not very polarisable, we are saying that the electron cloud of the atom cannot be easily distorted to give an uneven electron distribution, such that there are partial charges induced on different sides of the atom. Such atoms usually have high electron density. A good example often mentioned is the trend in polarisability in the halogens: Fluorine is the least polarisable while iodine is the most polarisable. This is due to the different sizes of the atom. Iodine having a larger and more diffuse electron cloud, allows for ease of electron movement within the electron cloud.
The effect of this is significant in bonding with hydrogen. When a halogen atom is bonded to hydrogen, there is evidently a partial positive charge on the hydrogen atom and a partial negative charge on the halogen atom. The polarisation of the iodine atom in hydrogen iodide is particularly significant due to its polarisable nature. The very uneven electron distribution of the iodine atom causes the hydrogen-iodine bond to be slightly unstable. On the contrary, since there is not so much of polarisation within the fluorine atom and thus, the hydrogen-fluorine bond is much more stable.
However, that is not the most important explanation for the bond strength to be as such. In fact, the "polarisability reason" as far less important, compared to the "orbital reason", which is that the fluorine atom has more defined orbitals used in bonding with hydrogen, compared to iodine. The interaction of more defined and electron-dense orbitals used in bonding gives rise to a greater accumulation of electron density in the bonding region. The attractive forces of the nuclei acting on the bonding electrons is thus stronger. Thus, there is a stronger bond.