(There is a misprint in the statement of the proposition. It should say $R=\operatorname{rad} P$.)
For the purposes of the proof, it is enough to consider the case when $U$ is indecomposable. The statement in green might be correct without that assumption, but $U$ being indecomposable is the only case needed for the proof of the proposition.
The authors want to prove $\begin{bmatrix} q \\ i \end{bmatrix}$ is left almost split. In other words, whenever $u \colon R \to U$ is not a section, one needs to show that $u$ factors through $\begin{bmatrix} q \\ i \end{bmatrix}$. As pointed out above, it suffices to consider the case when $U$ is indecomposable.
If $u \colon R \to U$ is a monomorphism, then, since $P$ is injective, there is a morphism $h \colon U \to P$ such that $i=hu$. $$R \xrightarrow{u} U \xrightarrow{h} P$$ But by Proposition 3.5, since $P$ is indecomposable projective, $i \colon R \to P$ is an irreducible morphism, meaning that in the factorization $i=hu$ either $u$ is a section or $h$ is a retraction. It is assumed $u$ is not a section, so $h$ must be a retraction. Since $U$ is indecomposable, it follows that $h$ is an isomorphism. So $u=h^{-1}i$ and the statement in green follows.