We for $\omega\in B(H)_*^+$ we have, $$|\omega(x-x_\alpha)|=|\omega(1^*(x-x_\alpha)|\leq \omega(1^* 1)^{1/2} \omega((x-x_\alpha)^*(x-x_\alpha))^{1/2}$$.
by applying Cauchy-Schwarz to the sesquilinear form $(x, y) \mapsto \omega(y^*x)$.
For a general $\omega\in B(H)_*$ we pick $x\in B(H)$ such that $\omega=\text{Tr}(x\cdot)$ and write $x$ as a linear combination of positive elements $x=\sum \alpha_i x_i$. Then $\omega=\sum \alpha_i\text{Tr}(x_i \cdot)$ and it easy to see that the functionals $\text{Tr}(x_i\cdot)$ are positive: write $x_i=z^*z$, then we have $\text{Tr}(x_i(y^*y))=\text{Tr}(z^*zy^*y)=\text{Tr}(yz^*zy^*)=\text{Tr}((zy^*)^*zy^*)\geq 0$ for all $y\in B(H)$. We conclude that $\omega$ is a linear combination of elements in $B(H)_*^+$.
Lemma: Let $(X_i)_{i\in I}$ be a collection of Banach spaces indexed by a set. Then there exists an isometric isomorphism
$$\bigoplus_{i\in I}^{\ell^\infty}(X_i)^*\cong\bigg(\bigoplus_{i\in I}^{\ell^1}X_i\bigg)^* $$
I leave the proof to you, but we can discuss the details if there is any problem.
By Sakai's theorem the predual is unique and thus the $\sigma$-weak topology is canonical, in the sense that it is an intrinsic object that is independent of the concrete representation of the von Neumann algebra. More specifically, if $M_i=\mathcal{B(H}_i)$, then
$$\bigoplus_{i\in I}^{\ell^\infty}M_i=\bigoplus_{i\in I}^{\ell^\infty}(M_{i*})^*\cong(\bigoplus_{i\in I}^{\ell^1}M_{i*})^*$$
where $A_*$ denotes the predual of the von Neumann algebra $A$. In other words, the predual of $M$ is the same as the $\ell^1$ direct sum of the preduals of the $M_i$, as you correctly guessed. You believe that this is not practical and you are wrong about that!
Fix $x=(x_i)_{i\in I}\in M$. We wish to find a net of elements such that only finitely many coordinates are non-zero that converges ultraweakly to our point $x$. Let $\Lambda$ be the set of finite subsets of $I$ and we direct $\Lambda$ by the usual inclusion, so $\Lambda$ becomes a directed set. We denote the elements of $\Lambda$ by $\lambda$, so $\lambda$ is simply a finite subset of $I$. For $\lambda\in\Lambda$ we define $x_\lambda=(x_\lambda^{(i)})_{i\in I}$ where $x_\lambda^{(i)}=0$ if $i\in I\setminus\lambda$ and $x_\lambda^{(i)}=x_i$ if $i\in\lambda$. So the net $(x_\lambda)_{\lambda\in\Lambda}$ lies in the subspace $N$ you have specified. We will show that $x_\lambda\to x$ ultraweakly.
We fix $\varepsilon>0$ and a functional $\Phi$ in the predual of $M$ and we want to find $\lambda_0\in\Lambda$ so that for any $\lambda\geq\lambda_0$ (here $\geq$ refers to the direction we gave to $\Lambda$ of course) we have that $|\Phi(x)-\Phi(x_\lambda)|<\varepsilon$.
Edit: Wait a minute; how am I fixing a functional in the predual, how does that make any sense? Well, recall that the predual of a von Neumann algebra $A$ is identified with the set of normal linear functionals on $A$ and therefore a net $(a_\mu)\subset A$ converges ultraweakly to $a\in A$ if and only if $\phi(a_\mu)\to\phi(a)$ for all normal linear functionals $\phi$. Now that we have cleared this out, let's go on with our proof:
By the identification of the preduals that we explained in the first paragraph, $\Phi$ is identified with a sequence $(\phi_i)_{i\in I}$ where $\phi_i\in M_{i*}$ such that $\sum_{i\in I}\|\phi_i\|<\infty$. Since the generalized series converges, we can find a finite subset $F\subset I$ so that $\sum_{i\in I\setminus F}\|\varphi_i\|<\frac{\varepsilon}{\|x\|}$. Set $\lambda_0=F$ and suppose that $\lambda\geq\lambda_0$, i.e. $\lambda\supset\lambda_0$. Then
$$|\Phi(x)-\Phi(x_\lambda)|=|\sum_{i\in I}\phi_i(x_i)-\sum_{i\in I}\phi_i(x_\lambda^{(i)})|=|\sum_{i\in I\setminus\lambda}\phi_i(x_i)|\leq\sum_{i\in I\setminus\lambda}|\phi_i(x_i)|\leq\sum_{i\in I\setminus F}|\phi_i(x_i)|\leq$$ $$\leq\|x\|\cdot\sum_{i\in I\setminus F}\|\phi_i\|<\varepsilon $$
and we are done.
Best Answer
Well, I guess Martin has already aptly (as always) answered this question but still let me expand on my previous comment, as suggested by the OP.
The first thing to notice is that the weak operator topology is not something intrinsic of a von Neumann algebra, as it depends on the underlying Hilbert space.
Some may say that von Neumann algebras are inextricably linked to the Hilbert space where they were born, and this is why people don't care at all about representing von Neumann algebras elsewhere, the prevailing attitude being to keep the algebra within its craddle $B(H)$.
Nevertheless, there is a very limited situation in which it might be useful to consider just one more representation, in addition to the defining one.
Given a von Neumann algebra $M\subseteq B(H)$, consider the representation of $M$ on $H\otimes \ell ^2$ given by $$ T\in M\mapsto T\otimes 1\in B(H\otimes \ell ^2). $$ Viewing $H\otimes \ell ^2$ as an infinite direct sum of copies of $H$, one has that $T\otimes 1$ is the (block) diagonal operator, with $T$ repeated down the diagonal infinitely many times.
$M\otimes 1$ is still weakly closed in $B(H\otimes \ell ^2)$, hence a von Neumann algebra, while the passage from $M$ to $M\otimes 1$ preserves all of the intrinsic properties of $M$. However, the weak operator topology may no longer be the same!
The reason is very simple. A vector $\xi $ in $H\otimes \ell ^2$ is actually a sequence $\xi =(\xi _n)_n$ of vectors, with each $\xi _n$ lying in $H$, and such that $\sum_n\|\xi _n\|^2<\infty $.
Observe that the vector state $\varphi _\xi $ defined by a vector $\xi $, as above, may be expressed, for each $T$ in $M$, by $$ \varphi _\xi (T) = \big \langle (T\otimes 1) \xi , \xi \big \rangle = \sum_n\big \langle T \xi _n, \xi _n\big \rangle . $$
Defining $u_n=\xi _n/\|\xi _n\|$, and $\alpha _n = \|\xi _n\|^2$, we then have that $$ \varphi _\xi (T) = \sum_n\alpha _n\big \langle T u_n, u_n\big \rangle = \sum_n\alpha _n\text{tr}(TS_n) = \text{tr}(TS), $$ where $S_n$ is the rank-one, trace-class operator $$ S_n(\xi ) = \big \langle \xi , u_n\big \rangle u_n, $$ and $S$ is the absolutely converging sum of the $\alpha _nS_n$.
Following along these lines one may prove that the vector states on $M\otimes 1$ in fact correspond to the normal states on $M$.
Summarizing, the passage from $M$ to $M\otimes 1$ turns the $\sigma $-weak operator topology into the weak operator topology.