Why does covalent bonding not break down if observer effect can be applied to atomic electrons?
I think part of your confusion stems from neglecting the fact that the electrons in a single molecule are all entangled with each other, so while an individual electron can occupy many states simultaneously, in each of those states it is entangled with all the other electrons such that the sum of states remains the same.
To start with a simple example, consider an atom of helium in its singlet ground state. Prior to measurement, neither electron has a defined spin. Both are a superposition of $+1/2$ and $-1/2$ states, but that does not mean that the helium atom can have $+1$ or $-1$ spin. The net spin is still 0. If we measure one electron's spin to be $+1/2$, the other will necessarily have $-1/2$, even though neither had defined spin prior to measurement.
We can extend this to multi-atomic molecules. Consider ground state methane. We can calculate delocalized molecular orbitals that collectively describe the charge density at any point in the molecule, with each of those orbitals being composed of a standard hydrogenic orbital of carbon (eg 2s or 2p) and athe 1s orbitals of all four hydrogen atoms. Alternatively, we can describe the same charge density distribution with four localized orbitals, each comprised of a single hydrogen 1s orbital and a hybrid sp3 orbital on carbon. Prior to making an observation, neither of these descriptions is correct, but neither is incorrect. Both correctly describe the charge density of the molecule, but both incorrectly assign electrons to specific states. Instead, the electrons are collectively in both states (and an infinite number of others that describe the same density). Because every contributing state correctly describes the charge density, there is no change in the covalent structure resulting from the superposition.