About Feynman's treatment of entropy

Feynman said in this chapter that if a system absorbs (rejects) an amount of heat $d Q$ at a temperature $T$, then we say the entropy of the system increased (decreased) by an amount $dS=d Q /T$.

If you re-read your link, you will see that Feynman did not say that. What you left out of your description is that Feynman explicitly specified that his equation applied to a reversible path. [The difference between $Q$ and $Q_{rev}$ is not a minor typo. Rather, it is crucial to understanding the issues you raise in your question.]

I.e., here is what he is saying:

$$dS=d Q_{\textbf{rev}} /T$$

[You only mention the reversibility constraint as applying to his equations for the efficiency of heat engines.]

Likewise, I'm afraid this is also incorrect, because it again leaves out the fact that Feynmann added the key constraint of reversibility:

He then said that when a system goes from state $a$ with temperature $T_{a}$ and volume $V_{a}$ to state $b$ with temperature $T_b$ and volume $V_b$, we can write the change of entropy as $$S_a-S_b=\int_{a}^{b} \frac{dQ}{T}.$$

Instead, Feynman's message is that we can calculate the entropy change associated with a process if we can connect the initial and final states by some reversible path (of which there are an infinite number), and integrate over that path, i.e.:

$$S_a-S_b=\int_{a}^{b} \frac{dQ_\textbf{rev}}{T}.$$

Entropy is a property of the system. It is a state function, and the change in the entropy of a system depends only on the difference between the intial and final states. The change in entropy is thus independent of the path through which that change is made; it doesn't matter whether the change is made reversibly or irreversibly—for a given initial and final state, the change in entropy will always be the same.

Heat flow, by contrast, is not a state function—i.e., it is not a property of the system. Rather (like work), it is a property of the path that connects changes in the system. Thus, if we use two different paths to connect an initial and final state, the heat and work flows along these paths can be different, even though the initial states for both paths are the same, and the final states also are the same.

Having said that, if we constrain a path in a very special way—namely, if we constrain the path to be reversible—we can calculate the change in entropy by integrating $\frac{dQ_\textbf{rev}}{T}$ along that path.