The sum of two extension is defined in the following way:
Let $\mathcal{B}$ and $\mathcal{A}$ be two separable C*-algebras with $\mathcal{B}$ stable and $\phi,\psi: \mathcal{A} \rightarrow \mathcal{M(B)/B}$ be two extensions of $\mathcal{B}$ by $\mathcal{A}$. Then the sum of extension is defined in the following way: $$(\phi \oplus \psi)(a)=\pi(S)\phi(a)\pi(S^*)+\pi(T)\psi(a)\pi(T^*)$$ where $S,T \in \mathcal{M(B)}$ are isometries such that $SS^*+TT^*=1_{\mathcal{M(B)}}$.
I do not follow the motivation why we define the sum in this way and what is the guarantee that we find two such isometries $S$ and $T$ such that $SS^*+TT^*=1_{\mathcal{M(B)}}$.
Can you refer me some notes or papers where I can find the details of this BDF sum?
Also I do not get the idea why we define the BDF equivalance of extensions as follows:
Two extensions $\phi$ and $\psi$ are equivalent if there exists $\rho,\sigma: \mathcal{A} \rightarrow \mathcal{M(B)/B}$ trivial extensions such that $$\phi \oplus \rho \sim_u \psi \oplus \sigma$$
Any reference of notes and paper will be very helpful. Thanks in advance.
Here's one way you can obtain the two isometries:
For any $C^*$-algebras $A$ and $B$, $M(A)\otimes M(B)$ always embeds unitally in $M(A\otimes B)$ (tensor products here are minimal). Thus if $B$ is stable, $M(B)\otimes\mathbb B$ embeds in $M(B\otimes\mathbb K)\cong M(B)$. So take two isometries $s,t\in\mathbb B$ such that $ss^*+tt^*=1_\mathbb B$, and put $S=1_{M(B)}\otimes s$, $T=1_{M(B)}\otimes t$.
For motivation, I would recommend consulting the original papers of Brown, Douglas, and Fillmore.