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.)
The general setup in topological spaces would be the following: given topological spaces $X,Y$ and a dense $X_0\subseteq X$ with $T:X_0\rightarrow Y$ continuous (for the subspace topology on $X_0$) we can extend $T$ by continuity to $X$. This is of course not true (exercise to the reader to find a counter-example for e.g. $X_0=(0,1), X=[0,1]$).
A true statement can be fashioned by using the theory of Uniform Spaces, Uniform Continuity and Completeness. I believe that then a general statement could be formulated and proved for topological vector spaces.
However, in this case, I think there are some missing details in Takesaki's proof, and I would prefer to give a different approach.
The way I would think about this problem is to use some Banach space theory. The general setup is to let $E,F$ be Banach spaces, let $E_0\subseteq E^*$ be a weak$^*$-dense subspace, and let $T_0:E_0\rightarrow F^*$ be bounded, linear, and weak$^*$-continuous. We claim we can extend $T_0$ to $T:E^*\rightarrow F^*$ a weak$^*$-continuous bounded linear map.
In fact, Takesaki proves something different. We can consider the Banach space adjoint (here I follow Takesaki's notation) ${}^t T_0:F^{**}\rightarrow E_0^*$, consider $F$ as a subspace of $F^{**}$ and so ask about ${}^t T_0(F) \subseteq E_0^*$.
Also each member of $E$ induces a functional on $E_0$, and by Hahn-Banach, weak$^*$-density of $E_0$ in $E^*$ implies that the resulting map $E\rightarrow E_0^*$ is at least injective. Takesaki shows that ${}^t T_0(F) \subseteq E$. So there is a linear map $S:F\rightarrow E$ making the following commute:
$$ \require{AMScd}\begin{CD} F^{**} @>{{}^t T_0}>> E_0^* \\
@AAA @AAA \\
F @>{S}>> E \end{CD} $$
Explicitly, given $\omega\in F$ and $x\in E_0\subseteq E^*$,
$$ \langle x, S(\omega) \rangle_{E^*,E} =
\langle {}^t T_0(\omega), x \rangle_{E_0^*, E_0}
= \langle T_0(x), \omega \rangle_{F^*,F}. $$
In general, I see no reason why $S$ need be bounded. Kaplansky density comes to the rescue, because $M_1 \otimes_{\min} M_2$ is a weak$^*$-dense $C^*$-subalgebra of $M_1 \overline\otimes M_2$, and so the unit ball is dense. Abstractly, this means we may add the additional hypothesis that the unit ball of $E_0$ is weak$^*$-dense in the unit ball of $E^*$. Equivalently (by Hahn-Banach again) $E_0$ norms $E$. So
\begin{align*}
\|S(\omega)\| &= \sup \{ |\langle x,S(\omega)\rangle| : x\in E_0, \|x\|\leq 1 \} \\
&= \sup \{ |\langle T_0(x), \omega \rangle| : x\in E_0, \|x\|\leq 1 \} \\
&\leq \|T_0\| \|\omega\|.
\end{align*}
Thus $S$ is bounded, with $\|S\| \leq \|T_0\|$.
Now set $T = {}^t S_0 : E^*\rightarrow F^*$ and it's an easy exercise to show that $T$ extends $T_0$; clearly $T$ is weak$^*$-continuous as it's a Banach space adjoint.
Best Answer
I follow the book of Effros+Ruan (which is a book, so not viewable online, but really is the nicest source I think). For any operator spaces $X,Y$ we can consider the operator space projective tensor product $\newcommand{\proten}{\widehat\otimes}X\proten Y$ whose dual satisfies $$ (X\proten Y)^* = CB(X,Y^*). $$ (To be precise, there is some completely isometric isomorphism here.) This is in Chapter 7, Corollary 7.1.5 to be precise.
The Fubini tensor product is also considered in Chapter 7, Theorem 7.2.3, which shows that $(X\proten Y)^* \cong X^* \bar\otimes_{\mathcal F} Y^*$ in general, so that $X^* \bar\otimes_{\mathcal F} Y^* \cong CB(X,Y^*)$.
So now apply this to a von Neumann algebra $M$ with predual $M_*$, and a dual operator space $X$ with predual $X_*$. Then $$ CB(M_*, X) = (M_* \proten X_*)^* = M \bar\otimes_{\mathcal F} X. $$ Again there are (canonical) isomorphisms involved here, but chasing them down will show that they match the isomorphism given in the original question. In particular, this includes the isomorphism $M \bar\otimes_{\mathcal F} X \cong X \bar\otimes_{\mathcal F} M$.
Here I used that $X$ is a dual space, and in Chapter 7 of Effros and Ruan, we need this, and a "dual realisation" of $X$ as a weak$^*$-closed subspace of $B(H)$. That is, the Original Question has a positive answer when $X \subseteq B(H)$ is weak$^*$-closed.
When $X$ is only assumed norm closed, we can use the definition of the Fubini tensor product given in the original question, though I am not aware of much study of this. However, let $\overline{X}^{w^*}$ be the weak$^*$-closure of $X$ in $B(H)$. Then by definition(s), $$ X \bar\otimes_{\mathcal F} M \subseteq \overline{X}^{w^*} \bar\otimes_{\mathcal F} M \cong CB(M_*, \overline{X}^{w^*}). $$ The isomorphism clearly identifies $X \bar\otimes_{\mathcal F} M$ with those CB maps $T:M_*\rightarrow \overline{X}^{w^*}$ which map into $X$, and by definition of what a CB map is, this is just the space $CB(M_*,X)$. Thus the original question is answered in the affirmative.