The interaction between the malicious Adversary and the Simulator and Real-Ideal' views
Without meaning any offense to the authors, this proof could perhaps have been written a bit more clearly; probably, since this is a fairly minor result, they didn't expend too much effort on it. Here is a longer but hopefully clearer rewording which should clear your terminological confusion.

Corrupted $P_1$

The real-world adversary $\mathcal A$ corrupting $P_1$ has input $x$ and auxiliary input $z$. It sends some value $x'$ to $P_2$ and outputs some value $x''$; both $x'$ and $x''$ are computed from $x$ and $z$. Since $P_2$ is honest, it computes and outputs $x' \wedge y$, for its received value $x'$ and input $y$, as instructed by the protocol; the output pair is $(x'', x' \wedge y)$.

The ideal-world adversary $\mathcal S$ (aka the simulator) again has input $x$ and auxiliary input $z$; it first computes $x'$ and $x''$ from $x$ and $z$ just like $\mathcal A$ does*. It then sends (or instructs $P_1$ to send) $x'$ to the trusted party; the honest $P_2$ sends $y$. The trusted party returns $\varepsilon$ (the empty string) to $\mathcal S$ and $x' \wedge y$ to $P_2$. $\mathcal S$ outputs $x''$ and $P_2$ outputs $x' \wedge y$; the simulation is perfect.

Corrupted $P_2$

The real-world adversary $\mathcal A$ corrupting $P_2$ has input $y$ and auxiliary input $z$. It receives $P_1$'s input $x$ and outputs some value $y'$ computed from $x$, $y$ and $z$. Since $P_1$ is honest, it outputs $\varepsilon$ as instructed by the protocol; the output pair is $(\varepsilon, y')$.

The simulator $\mathcal S$ again has input $y$ and auxiliary input $z$; it sends $1$ to the trusted party, and the honest $P_1$ sends $x$. The trusted party returns $\varepsilon$ to $P_1$ and $x \wedge 1 = x$ to $\mathcal S$. The honest $P_1$ outputs $\varepsilon$, and $\mathcal S$ computes and outputs $y'$ just like $\mathcal A$ does; the simulation is perfect.

* This is basically what they mean by "$\mathcal S$ receives [...] from $\mathcal A$." You could consider that $\mathcal S$ is given black-box or oracle access to $\mathcal A$: it sends $\mathcal A$ the values $x$ and $z$ and $\mathcal A$ returns $x'$ and $x''$ like it does in the real world. It's a bit awkward, however, because such black-box/oracle access has not been defined at this point in the book, and in particular does not appear in the security definitions.