This is the Kraus Theorem. But, the map $\mathcal{M}$ is the dual map of a map $\Phi : \mathfrak{T}(\mathcal{H}) \to \mathfrak{T}(\mathcal{H})$ where $\mathfrak{T}(\mathcal{H})$ is the banach space of trace-class operators. Then from the theorem above, is it possible to obtain an expression analogous to (6.14) for $\Phi$? I am very despeate and curious... Could anyone please help me?
Let $\Phi$ be a linear and completely positive map on $\frak T(\mathcal H)$ where $\mathcal H$ is a separable Hilbert space. Then its dual map $\Phi^*$ acting on $\mathcal B(\mathcal H)$ is linear and completely positive (the equivalence of complete positivity of $\Phi$ and $\Phi^*$ is well known) and the duality relation implies that $\Phi^*$ is ultraweakly (here: $\sigma$-weakly) continous. Now we can apply Theorem 6.23 to $\Phi^*$ to get
$$ \operatorname{tr}(\Phi(A)B)=\operatorname{tr}(A\Phi^*(B))\overset{(6.14)}=\operatorname{tr}\Big(A\sum_n M_n^*BM_n\Big)=\sum_n\operatorname{tr}(AM_n^*BM_n)\\=\sum_n\operatorname{tr}(M_nAM_n^*B)=\operatorname{tr}\Big(\sum_n M_nAM_n^*B\Big) $$
for all $A\in\frak T(\mathcal H), B\in\mathcal B(\mathcal H)$ using that the trace is linear, continous and cyclic. Thus every such $\Phi$ has a representation
$$ \Phi(A)=\sum_n M_nAM_n^*\tag{1} $$
for all $A\in\frak T(\mathcal H)$ where $\sum_n M_n^*M_n$ converges strongly and (1) even converges with respect to the trace norm (by Proposition 6.3 of that Attal chapter you quoted in your question).