Either can work correctly if designed properly. If you have a dumb rectifier supply feeding a 7805, then all the rectifier part needs to do is guarantee the minimum input voltage to the 7805 is met.
The problem is that such a power supply only charges up the input cap at the line cycle peaks, then the 7805 will drain it between the peaks. This means the cap needs to be big enough to still supply the minimum 7805 input voltage at the worst case current drain for the maximum time between the peaks.
The advantage of a full wave rectifier is that both the positive and negative peaks are used. This means the cap is charged up twice as often. Since the maximum time since the last peak is less, the cap can be less to support the same maximum current draw. The downside of a full wave rectifier is that it takes 4 diodes instead of 1, and one more diode drop of voltage is lost. Diodes are cheap and small, so most of the time a full wave rectifier makes more sense. Another way to make a full wave rectifier is with a center tapped transformer secondary. The center is connected to ground and there is one diode from each end to the raw positive supply. This full wave rectifies with only one diode drop in the path, but requires a heavier and more expensive transformer.
A advantage of a half wave rectifier is that one side of the AC input can be directly connected to the same ground as the DC output. That doesn't matter when the AC input is a transformer secondary, but it can be a issue if the AC is already ground-referenced.
There's nothing particularly mystifying about this circuit.
The left-most op-amp circuit is an inverting unity-gain precision half-wave rectifier. The output is zero during the positive half-cycle of the input and is the inverted input during the negative half-cycle.
The right-most op-amp circuit is a non-inverting unity gain amplifier so the output equals the input which is just the output of the left-most circuit.
At the output node, we have by voltage division and superposition:
$$v_{fw} = \frac{1}{3} v_7 + \frac{2}{3} v_{hw}$$
During the positive half-cycle, the right-most term is zero so the output is just \$v_{fw} = \frac{1}{3} v_7\$.
During the negative half-cycle, the right-most term is \$-\frac{2}{3} v_7\$ and so, the sum of the two terms is \$v_{fw} = -\frac{1}{3} v_7\$.
Thus, the output is an attenuated full-wave rectified version of the input:
$$v_{fw} = \frac{1}{3}|v_7|$$
Best Answer
In each case, look at what happens when the first stage cuts off.
In the simpler circuit, the first opamp's output hits the negative supply rail and stays there until the input goes negative again. Then it slowly comes out of saturation and ramps up at its slew rate until finally the diode starts conducting again.
In the second circuit, D2 keeps it out of saturation and the output only has to transition across 2 diode drops to start conducting again.
This will impact its accuracy for high frequency inputs.
If you're only rectifying low frequency signals, or the opamp behaves well in saturation and has a high slew rate, the first circuit is OK, but if you need accuracy on high frequency signals with a low cost opamp, the second may meet your goals more economically.