If you're working at the level of Hilbert spaces, I think the more usual procedure nowadays is to fix a distinguished unit vector $u_i$ in each Hilbert space $H_i$, then define the tensor product to be the Hilbert space generated by all elementary tensors $\bigotimes v_i$ such that $v_i \in H_i$ for all $i$ and $v_i = u_i$ for all but finitely many $i$. This is a direct limit of the tensor products over finite sets of indices in a fairly obvious way. In physical models the $u_i$ would typically be ground states. This definition seems more usable than the old "full" infinite tensor product defined by von Neumann which is typically nonseparable. I give details in Section 2.5 of my book Mathematical Quantization.

It seems that the infinite tensor product defined by Nakagami in general will not act faithfully on the Hilbert space defined above. I am not sure what to make of this fact. An analogous construction of an infinite tensor product of von Neumann algebras can be given on my Hilbert space tensor product. (Say each $M_i$ acts on $H_i$, then each $M_i$ acts on $\bigotimes H_i$ and we let $\bigotimes M_i$ be the von Neumann algebra generated by all the $M_i$ within $B(\bigotimes H_i)$. But since my Hilbert space is just a piece of the full tensor product, I expect my von Neumann algebra tensor product to be a quotient of Nakagami's. So this gets kind of interesting!

I suppose the predual of $\bigotimes M_i$ as I am defining it will be a direct limit of the maximal tensor products of finite sets of preduals of the $M_i$, with the embedding maps given by tensoring with the vector states coming from the $u_i$.

For your last question, in my setup this certainly works if $\omega$ is the vector state coming from $u_i$, and in Nakagami's definition you'll be okay as long as $\omega$ is any vector state. So it seems like you want to be sure $M$ is in standard form to begin with, so that every normal state is a vector state, and then Nakagami's construction should do what you want. (Otherwise I imagine it doesn't.)

If you're working at the level of Hilbert spaces, I think the more usual procedure nowadays is to fix a distinguished unit vector $u_i$ in each Hilbert space $H_i$, then define the tensor product to be the Hilbert space generated by all elementary tensors $\bigotimes v_i$ such that $v_i \in H_i$ for all $i$ and $v_i = u_i$ for all but finitely many $i$. This is a direct limit of the tensor products over finite sets of indices in a fairly obvious way. In physical models the $u_i$ would typically be ground states. This definition seems more usable than the old "full" infinite tensor product defined by von Neumann which is typically nonseparable. I give details in Section 2.5 of my book Mathematical Quantization.

It seems that the infinite tensor product defined by Nakagami in general will not act faithfully on the Hilbert space defined above. I am not sure what to make of this fact. An analogous construction of an infinite tensor product of von Neumann algebras can be given on my Hilbert space tensor product. (Say each $M_i$ acts on $H_i$, then each $M_i$ acts on $\bigotimes H_i$ and we let $\bigotimes M_i$ be the von Neumann algebra generated by all the $M_i$ within $B(\bigotimes H_i)$. But since my Hilbert space is just a piece of the full tensor product, I expect my von Neumann algebra tensor product to be contained in Nakagami's. So this gets kind of interesting!

I suppose the predual of $\bigotimes M_i$ as I am defining it will be a direct limit of the maximal tensor products of finite sets of preduals of the $M_i$, with the embedding maps given by tensoring with the vector states coming from the $u_i$.

For your last question, in my setup this certainly works if $\omega$ is the vector state coming from $u_i$, and in Nakagami's definition you'll be okay as long as $\omega$ is any vector state. So it seems like you want to be sure $M$ is in standard form to begin with, so that every normal state is a vector state, and then Nakagami's construction should do what you want. (Otherwise I imagine it doesn't.)