In Rordam's book, there is a lemma before giving the definition of index map.
I am quite confused about two places in the proof of conclusion (i).
In the proof, it mentiones that $\tilde{\varphi}$ is injective, then there exists $p\in M_{2n}(\tilde{I})$ such that $\tilde{\varphi}(p)=vdiag(1_n,0)v^*$, how to conclude the above conclusion?
$s$ is a map from $M_{2n}(\tilde{I})$ to $M_{2n}(\tilde{I})$, but $diag(1_n,0)\in M_{2n}(\tilde{B})$, how to show that $s(p)=diag(1_n,0)$?
Lemma 4.3.1 really is crucial here (and as mentioned in the text, the proofs are straightforward). Also for your question (1), it is "$\tilde{\phi}$ is injective and there exists ...". The fact that there is an element $p$ follows from Lemma 4.3.1 $(ii)$ since $$\tilde{\psi}\left(v\begin{pmatrix}1_n & 0 \\ 0 & 0\end{pmatrix}v^*\right) = \begin{pmatrix}1_n & 0 \\ 0 & 0\end{pmatrix} = s\left(\begin{pmatrix}1_n & 0 \\ 0 & 0\end{pmatrix}\right), $$ so that $$v\begin{pmatrix}1_n & 0 \\ 0 & 0\end{pmatrix}v^* $$ is in the image of $\tilde{\phi}$ (really $\tilde{\phi}_{2n}$); i.e., there exists $p \in M_{2n}(\tilde{I})$ such that $\tilde{\phi}(p) = v\begin{pmatrix}1_n & 0 \\ 0 & 0\end{pmatrix}v^*$. The fact that $p$ is a projection follows from the injectivity of $\tilde{\phi}$ ($\tilde{\phi}$ is injective by Lemma 4.3.1 (i)). I believe this is the only place that the injectivity of $\tilde{\phi}$ is used with regard to your questions, but please correct me if I'm wrong.
Now to see why $s(p)$ is the desired matrix, the scalar map (as in 4.2.1) is natural in the sense that it commutes with *-homomorphisms (extended to unitizations), so $\tilde{\phi}\circ s = s\circ\tilde{\phi}$ and $\tilde{\psi}\circ s = s \circ \tilde{\psi}$ (where $s$ has its proper domain and codomain in each respective case). In particular we have that $s \circ (\tilde{\psi}\circ \tilde{\phi}) = (\tilde{\psi}\circ\tilde{\phi})\circ s$. So since $$\begin{pmatrix}1_n & 0 \\ 0 & 0\end{pmatrix} = s\left(\begin{pmatrix}1_n & 0 \\ 0 & 0\end{pmatrix}\right), $$ it follows that $$ \tilde{\psi}\circ\tilde{\phi}(s(p)) = \begin{pmatrix}1_n & 0 \\ 0 & 0\end{pmatrix}. $$ Now these maps between unitizations are "injective on scalars" in a sense. By this I mean you can recover the scalar part of an element in the unitization from its image: if $\tilde{f}: \tilde{C} \to \tilde{D}$ is the unitization of a *-hom $f: C \to D$, then $\tilde{f}(x) = y + \alpha \tilde{1}, y \in D, \alpha \in \mathbb{C}$, implies that $x = a + \alpha \tilde{1}$ for some $a \in C$. So if we are only considering scalars, we can completely recover them. Consequently since $\tilde{\psi}\circ\tilde{\phi}(s(p)) = \text{diag}(1_n,0)$ (where $s(p)$ and $\text{diag}(1_n,0)$ are both scalars), it follows that $$ s(p) = \begin{pmatrix}1_n & 0 \\ 0 & 0\end{pmatrix}. $$ The paragraph above still makes sense when you take your induced maps between matrix algebras, which is what is really happening here. I just suppressed the subscripts as is done in the text.