I think the way to understand this is to go back to his treatment of the LSZ reduction formula. To reiterate the basics, what we want to compute is the amplitude for our state $|\psi(t)\rangle$ to be in some state $|f\rangle$ at some time $t$, i.e. $$\langle f|\psi(t)\rangle=\langle f|S(t, t_0)|\psi(t_0)\rangle,$$ where $S(t, t_0)$ is the time-evolution operator. To make life easy, we assume that the initial and final states are far removed from interactions in the far past and far future, respectively, and are multi-particle states of definite momentum. Let's try to construct these states for $2\to n$ scattering. Remember that we are working in the Schrodinger picture here (hence the time-evolution operator), so operators are time-independent. Our initial and final states are (note the differences to Schwartz)
\begin{align*}
|\psi(-\infty)\rangle&=\sqrt{2\omega_1}\sqrt{2\omega_2}\,a_{p_1}^\dagger a_{p_1}^\dagger|\Omega(-\infty)\rangle=|p_1,p_2\rangle\\|f\rangle&=\sqrt{2\omega_3}\cdots\sqrt{2\omega_n}\,a_{p_3}^\dagger\cdots a_{p_n}^\dagger|\Omega(\infty)\rangle=|p_3,\dots,p_n\rangle
\end{align*}
($|\psi(-\infty)\rangle\equiv|i\rangle$), giving the amplitude $$\langle p_3,\dots,p_n|S(\infty, -\infty)|p_1,p_2\rangle=2^{n/2}\sqrt{\omega_1\cdots\omega_n}\langle\Omega(\infty)|\,a_{p_3}\cdots a_{p_n}S(\infty,-\infty)a_{p_1}^\dagger a_{p_2}^\dagger|\Omega(-\infty)\rangle.$$
The claim here is that, while the vacuum at asymptotic times is still technically the interacting vacuum, we are far enough removed from interactions that acting with the free-theory creation operators creates genuine single-particle states of definite momenta. Since $a_p=a_p(t_0)$, the annihilation operator at the reference time $t_0$ (when the Schrodinger and Heisenberg pictures are equal), it follows that $\boldsymbol{a_p(t_0)}$ annihilates $\boldsymbol{|\Omega(\pm\infty)\rangle}$.
Now let's move to the Heisenberg picture. In terms of the time evolved $a$'s, $$a_p=a_p(t_0)=S^\dagger(t_0,\infty)a_p(\infty)S(t_0,\infty).$$ Applying this to the vacuum then gives \begin{align*}a_p(t_0)|\Omega(\infty)\rangle&=S^\dagger(t_0,\infty)a_p(\infty)S(t_0,\infty)|\Omega(\infty)\rangle\\&=S^\dagger(t_0,\infty)a_p(\infty)|\Omega(t_0)\rangle,\end{align*} and similarly for $-\infty$. So we find that $\boldsymbol{a_p(\pm\infty)}$ annihilates $\boldsymbol{|\Omega(t_0)\rangle}$, the fixed vacuum of the Heisenberg picture that Schwartz simply refers to as $|\Omega\rangle$.
The above should answer questions 2 and 3 and make sense of the operations that Schwartz carries out. As for question 1, I have to admit I'm also somewhat confused by what he's written. But the point of this activity is that we want to express the correlation function in terms of the free fields and free vacuum because we know what they look like and their action on one another is simple. This allows for the eventual contraction into Feynman propagators, at which point the rest of the LSZ calculation is relatively straightforward.
