$\sigma$-weak topology versus weak operator topology The reference text for this question is: 
Pedersen, Analysis Now, GTM 118.
The $\sigma$-weak topology on $B(H)$ (the bounded linear operators on a Hilbert space $H$) is the weak$^*$-topology on $B(H)$ seen as the dual of the Banach algebra $B_1(H)$ ($T\in B_1(H)$ if $T$ is a compact operator such that its norm
in $B_1(H)$ is $tr(|T|)<\infty$ - see Pedersen 4.6.10, 3.4.6, 3.4.12, 3.4.13).
The weak operator topology on $B(H)$ is the one generated by the semi norms
$|(Tx,y)|$ as $x,y$ range in $H$ and $(\cdot,\cdot)$ is the inner product on $H$
- see Pedersen 4.6.1.
It is shown in Proposition 4.6.14 that the weak operator topology and the 
$\sigma$-weak topology overlap on the ball $B(H,n)$ of operators $T\in B(H)$ of norm at most $n$.
This gives that any $\sigma$-weakly continuous functional $\phi$ when restricted to $B(H,1)$ is continuous with respect to the weak operator topology.
Now propositions 4.6.11 and 4.6.4 give the following characterizations of continuous functionals in these topologies:
4.6.11: a functional $\phi:B(H)\to\mathbb{C}$ is $\sigma$-weakly continuous if and only if there exists sequences $(x_n)_{n\in\mathbb{N}}$ and $(y_n)_{n\in\mathbb{N}}$ of elements of $H$ such that
$\sum_{n\in\mathbb{N}}||x_n||+\sum_{n\in\mathbb{N}}||y_n||<\infty$ and
$\phi(S)=\sum_{n\in\mathbb{N}}(S(x_n),y_n)$.
4.6.4: a functional $\phi:B(H)\to\mathbb{C}$ is continuous with respect to the weak operator topology if and only if there exists sequences 
$(x_0,\dots, x_n)$ and $(y_0,\dots,y_n)$ of elements of $H$ such that
$\phi(S)=\sum_{j=0,\dots, n}(S(x_j),y_j)$.
I can follow the proof of proposition 4.6.14, but I don't understand how this
translates at the level of these different characterizations, i.e.:
Assume $\phi(S)=\sum_{n\in\mathbb{N}}(S(x_n),y_n)$ is $\sigma$-weakly continuous, what are the vectors $(z_0,\dots, z_n)$ and $(w_0,\dots,w_n)$ such that $\phi(S)=\sum_{i\leq n}(S(z_i),w_i)$ for all operators $S$ of norm at most 1?
How can it happen that a $\sigma$-weak continuous $\phi$ functional is such that
$\phi\restriction B(H,n)=\psi_n$ with $\psi_n$ continuos for the weak operator topology but for all $n$ $\phi\neq\psi_n$?
Some examples would be illuminating.
Thanks in advance
 A: For your first question, I think you are misunderstanding what the theorems say. When you restrict your $\sigma$-weakly continuous functional to the unit ball, you don't get a wot-continuous functional on the whole space: so 4.6.4 does not apply. 
For your second question, here is an example: fix an orthonormal basis $\{e_n\}$ and let 
$$
\phi(S)=\sum_n\frac1{n^2}\langle Se_n,e_n\rangle.
$$
This is clearly $\sigma$-weakly continuous by 4.6.11. And by 4.6.14 (I'm using your numbers, I don't have the book at hand) it is weakly continuous on bounded sets. But it is not weakly continuous: here is an example I posted a few days ago. I'm not sure if you are aware, but the continuity does not fail on sequences (a weakly convergent sequence is bounded) so we need nets. 
Let $\mathcal F=\{F\subset H:\ F \text{ is a finite-dimensional subspace} \}$, ordered by inclusion. We construct a net of operators indexed by $\mathcal F$ as follows:
$$
T_F=\frac1{\phi(P_{F^\perp})}\,P_{F^\perp},
$$
where $P_{F^\perp}$ is the orthogonal projection onto $F^\perp$. 
Then, for any $x\in H$, if we move far enough along the net we will have $x\in F$, so $T_Fx=0$, and then
$$
\langle T_Fx,y\rangle=0
$$
for any $x,y$. That is, $T_F\to0$ weakly.
But, for any $F$,
$$
\phi(T_F)=\frac{\phi(P_{F^\perp})}{\phi(P_{F^\perp})}=1.
$$
So $\{\phi (T_F)\}$ does not converge to $\phi (0)=0$, showing that $\phi $ is not weakly continuous. 
