Nucleophilic acyl substitution leaving group
Indeed, there is no reason why the protonated alcohol should not be eliminated upon the lone pair rebound. It probably happens more often than not. However, it does not lead to an observable product: it just goes back to the reactants. Thus, we can write an equilibrium arrow on the first step.

$$\ce{R-C(=O)-Cl + R'-OH &lt;=&gt; R-C(-O^-)(-Cl)-OH+-R'}\tag{1}$$

Once the chloride has been kicked out, there is no way back as the ester is a much weaker electrophile than the acid chloride (and chloride a weaker nucleophile than the alcohol). Thus, the reason why we only care about the lone pair rebound eliminating chloride is because only that is productive.

$$\ce{R-C(-O^-)(-Cl)-OH+-R' -&gt; R-C(=O)-OH+-R' + Cl-}\tag{2}$$

These arrows are also given in your source like that; if everything up to a single step is given as equilibria while that one step is given as non-reversible, that should be sufficient driving force to let a reaction happen.

At which point in time the proton is eliminated is somewhat up to debate, of little relevance for the overall reaction and dependent on what is in solution. If just a pure acid chloride and a pure alcohol are mixed the only potential bases that could accept the proton of the tetrahedral intermediate are other acid chloride molecules or other alcohols — or the tetrahedral intermediate’s $\ce{O-}$ atom. All of these are very weak bases and I see no preference for the proton to go anywhere. After we have eliminated chloride, that is added to the set of potential bases. It is also not a strong base; maybe even a weaker base than some of the other options. The big difference is that it is volatile and can diffuse out of the reaction mixture. Thus, with no other bases added, I would actually prefer this sequence of steps.

Once you add a base, I would also prefer deprotonation before chloride is eliminated.