The sample space is $\Omega$ with $\omega = (\omega_0, \omega_1, \ldots) \in \Omega$ an infinite sequence of a set $S$. So the measure space is $(S^{\mathbb{N}}, \mathcal{S}^{\mathbb{N}})$ where $\mathcal{S}$ is the $\sigma$-field of $S$.
The shift operator $\theta_n:\Omega \to \Omega$ acts like $(\theta_n \omega)_m = \theta_{n+m}$ (i.e. applying the shift $\theta_n$ to $\omega$ removes the first $n$ elements of the sequence.
Durrett's book says the Markov property is $E(Y\circ \theta_n|\mathcal{F}_n) = E_{X_n}Y$ when $Y \in \mathcal{F}$ and is bounded. He suggests thinking of $Y = h(\omega_0, \omega_1, \ldots)$ where $h$ is a bounded function in which case I think the Markov property can be written as $E(h(\omega_n, \omega_{n+1}, \ldots)|\mathcal{F}_n) = E_{X_n}(h(\omega_0, \omega_1, \ldots))$ which I do not really understand.
I know that $\omega = (\omega_0, \omega_1, \ldots)$ is a sequence and it is is in the sample space and it is meant to represent a Markov chain, right? But when you look at the "normal" Markov property that $P(X_{n+1} = j | X_0, \ldots, X_n) = P(X_{n+1} = j | X_n)$ and compare it to the property above, the right hand side includes all elements in the chain! I know the expectation is taken wrt X_n only but the $h(\omega_0, \ldots)$ includes all the chain.
Or is this $\omega$ different to the $X_n$?