Judging from the information you provided, it feels that the source is wrong. I don't know where your picture comes from. If it comes from a domestic newspaper, it is probably wrong. If it is a textbook, it is not easy to make mistakes. You can ask another clever teacher. Let me say my opinion for your reference:

The middle picture should be negative half-wave rectification, and the upper picture should be full-wave rectification.

You can feel that the waveforms are the same, which means that your input analysis of the negative half cycle is correct, so I won't go into details here. I only explain the difference between positive and negative Tuesdays. Using the instantaneous polarity method,

Alignment: When the Vo input is positive, the output of the operational amplifier is negative. At this point, D 1 is on and D2 is off. This is a transient state, which can be regarded as a short circuit at D 1 and an open circuit at D2. Immediately turn to the steady state, that is, the operational amplifier forms negative feedback, the virtual short circuit condition is established, and the positive and negative inputs of the operational amplifier are short-circuited to ground-the output is also zero. This means that the positive half cycle is "flat-topped", so I say that the middle picture is negative wave rectification.

Compare the above figure: When Vo input is positive, the output of operational amplifier is negative. At this time, D is turned off, and the output of the operational amplifier is considered as an open circuit, so Vo and Vi are directly connected through R2 and R 1, and their potentials are equal, that is to say, this circuit does not process the positive half-cycle input, and directly passes through the original waveform, so I say that the above figure is full-wave rectification. (At this time, the operational amplifier does have negative saturation output, and the source is correct. )

It seems that either the source has reversed the serial numbers of the two pictures or the landlord has read them backwards.

About the following figure: At least it can be understood as a first-order active filter circuit, and the operational amplifier does play the role of a buffer in the back. In order to convert the output impedance of front-end first-order passive filtering, this is the most commonly used strategy for analog active filtering. As for C3, it may be for frequency compensation of operational amplifier.

That's all I thought of. I hope it helps you.