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I'm looking at the following lecture notes where we start with the circuit below for some state $vertpsirangle_L$ that picks up an error to become $Evertpsirangle_L$

It is later claimed in the notes that the syndrome extraction part of the circuit can be represented by the following operation on $Evertpsirangle_L$.

$$E|psirangle_L|0rangle_A rightarrow frac12left(I_1 I_2+Z_1 Z_2right) E|psirangle_L|0rangle_A+frac12left(I_1I_2-Z_1 Z_2right) E|psirangle_L|1rangle_A$$

How does one see this? I can write the Hadamard and the control $Z_1Z_2$ gates as 8x8 matrices but this seems like a tedious way to do it. The alternative is to express the control $Z_1Z_2$ gates using something like this answer. However, I was unable to do it this way either.

So the question is - how do I see that the following line is true just by looking at the circuit?

$$E|psirangle_L|0rangle_A rightarrow frac12left(I_1 I_2+Z_1 Z_2right) E|psirangle_L|0rangle_A+frac12left(I_1I_2-Z_1 Z_2right) E|psirangle_L|1rangle_A$$

Let's represent controlled $Z_1Z_2$ gate in the projector formalism, as described in this answer:

$$C_AZ_1Z_2 = |0ranglelangle0|_A I_1I_2 + |1ranglelangle1|_A Z_1Z_2 $$

This just tells you to apply identity gates to qubits 1 and 2 if the ancilla is in the $|0rangle$ state and to apply Z gates to qubits 1 and 2 if the ancilla is in the $|1rangle$ state - which is the definition of the controlled gate.

Now let's apply this to the state $colorblue_AE|psirangle_L$ (this is the state of the system after the first Hadamard gate of syndrome extraction):

$$C_AZ_1Z_2 big( colorblue_AE|psirangle_L big) = big( colorblue0ranglelangle0_A I_1I_2 + colorblue_A Z_1Z_2 big) bigg( frac1sqrt2colorblue(_AE|psirangle_L bigg) = $$

$$= frac1sqrt2 big( colorblue0rangle_A otimes I_1I_2 E|psirangle_L + colorblue_A otimes Z_1Z_2 E|psirangle_L big)$$

Finally, apply the last Hadamard gate to the ancilla; after that the state of the system becomes

$$frac1sqrt2 big( colorblue_A otimes I_1I_2 E|psirangle_L + colorblue_A otimes Z_1Z_2 E|psirangle_L big) = $$

$$= frac12 big( colorblue(_A otimes I_1I_2 E|psirangle_L + colorblue1rangle)_A otimes Z_1Z_2 E|psirangle_L big) = $$

(after reordering the terms and grouping same ancilla states together)

$$= frac12 colorblue0rangle_A otimes left(I_1 I_2+Z_1 Z_2right) E|psirangle_L + frac12 colorblue_A otimes left(I_1I_2-Z_1 Z_2right) E|psirangle_L$$

which is exactly the state you need to get.