What exactly does the saturation region mean in terms of the physical charge distribution of the pnp/npn junction? Same for ON state.

Referring to the switch in Eggleston Fig 4.7, the ON state corresponds to the case when $V_1$ (and $I_b$) are large enough that the intersection of the straight line and the $I_c$ vs $V_{ce}$ curve for the appropriate $I_b$ lies in the left region where all the curves are mashed up against each other and $I_{c}$ is rapidly increasing with $V_{ce}$. This is the ``saturation region''. Physically, inside the transistor, the collector-base junction is no longer reverse-biased and there is no longer a large depletion zone at the collector-base junction (in fact both junctions are forward-biased). You get a relatively large $I_c$ for a given $V_{ce}$.

(Why is this called the ``saturation region''? I think it's because the relevant region is close to the plateau, where $I_c$ is just starting to saturate, i.e. reach a maximum value for a given $I_b$. It actually think it might make more sense to call the plateau region the saturation region, because there the current is truly saturated... but I did not make up this nomenclature. ``Linear active region'' for the plateau is not a bad name though. The plateau region is also known as the ``forward-active region'', also an okay name. Maybe because the $I_c$ value in this ``almost-saturated'' region is pretty independent of $I_b$ favors just calling it the ``saturation region''.)

In the OFF state, the pn junctions inside the transistor are fully depleted zones; electrons cannot easily jump the potential difference and create current from collector to emitter. The transistor can be considered an open circuit between emitter and collector in this state.

Current direction is opposite for a pnp transistor; otherwise the concepts are the same.