Why No Change in Polarizability in Asymmetric Stretch?
Please explain using plain english (without any rules or tables) why is that in asymmetric stretching like the following, there is no change in the size, shape or orientation of the electron cloud that surrounds the molecule?

There is a change. The size and shape of the electron cloud does change for the individual bonds affected by the asymmetric stretch. Polarizability relates to the ease with which electrons can be moved from their equilibrium position.  Electrons in weaker (longer) bonds are more easily displaced than electrons in stronger (shorter) bonds.
From your diagram, I believe that you are discussing the case of a linear $\ce{AB2}$ molecule. Picture the asymmetric stretch in our $\ce{AB2}$ molecule with the two $\ce{B}$ atoms remaining stationary and the $\ce{A}$ atom oscillating back and forth. Simultaneously one $\ce{A-B}$ bond becomes longer and the other shorter.  One becomes more polarizable, the other less polarizable.  In the linear $\ce{AB2}$case these changes in polarizability are equal and opposite, hence for the molecule overall they cancel out and there is no change in overall polarizability (for example, see here).

In order for a molecule to be Raman active, there must be a change in the polarizability

Yes, but it must be a change in polarizability for the entire molecule, not just an individual bond. Further, keep in mind the  Principle of Mutual Exclusion. It states that, for centrosymmetric molecules (molecules with a center of symmetry, your linear $\ce{AB2}$ is an example), vibrations that are IR active are Raman inactive, and vice versa.
The asymmetric stretch in such a molecule changes the dipole moment in the molecule from zero to non-zero; hence the asymmetric stretch is IR active. Therefor by the Principle of Mutual Exclusion, the asymmetric stretch must be Raman inactive.  Even though the asymmetric stretch changes the polarizabilty of some bonds in the molecule, the polarizability of the whole molecule does not change (as explained above) and the asymmetric stretch is Raman inactive as predicted.
The links I posted in the comments show how these concepts can be applied in the linear $\ce{AB2}$ examples of $\ce{CO2}$ and $\ce{XeF2}$.