directed diagrams in accessible categroies I have a problem with understanding the choice of $i_0$ and $i_1$ in the snippet below.
I believe that they cannot be completely arbitrary and that $\lambda^+$-directedness of $I$ must be used in some way. I have understood most parts of this (nice and difficult) paper, but I do not have a definite and simple answer for this point. I think that also
the morphism $d_{i_0,i_1}$ is the key here. Some intuition for the choice
will be welcome. I apologize to all who do not like snippets in posts, but this was the most easy way to report the problem.

 A: For finding $i_0$ we do indeed need the property that $I$ is $\lambda^+$-directed. By assumption $A$ is $\lambda^+$-presentable and $A$ is a $\lambda^+$-directed colimit with cocone $(d_i: M \to A)_{i \in I}$. So the identity arrow $Id_A: A \to A$ must factor through the cocone. That means precisely that there is some $i_0 \in I$ such $Id_A$ factorises through $d_{i_0}$, and that just means that there is some $f: A \to M$ such that $Id_A = d_{i_0} f$.
From that last equality it follows that $d_{i_0} (f d_{i_0}) = d_{i_0} Id_M$. We can now use this to find $i_1$. From earlier we know that $M$ is $\omega$-presentable (see Lemma 4), so in particular $M$ is $\lambda^+$-presentable. Thus the arrow $d_{i_0} (f d_{i_0}): M \to A$ must factor through the diagram $(d_i: M \to A)_{i \in I}$. Of course, we already have a factorisation for this, two even: $M \xrightarrow{f d_{i_0}} M \xrightarrow{d_{i_0}} A$ and $M \xrightarrow{Id_M} M \xrightarrow{d_{i_0}} A$. Now it becomes important that we do not just get any factorisation, but an essentially unique factorisation (I recalled this definition at the bottom of this answer). This means precisely that we can find some $i_1 \in I$ with $i_0 \leq i_1$ such that $d_{i_0, i_1} (f d_{i_0}) = d_{i_0, i_1} Id_M$, where $d_{i_0, i_1}$ is the arrow in the diagram corresponding to the arrow $i_0 \leq i_1$ in the indexing category.
Now the argument can be finished as in the paper, noting that $d_{i_0, i_1}$ is an arrow $M \to M$ and is thus in particular in $(M, \omega)\text{-}\mathbf{Set}$, so part (a) applies to this arrow.

Let me recall the definition of essentially unique. The notation here has nothing to do with the above answer. Let $C = \operatorname{colim}_{i \in I} D_i$ be a $\lambda$-directed  colimit, with coprojections $d_i: D_i \to C$. Then if $A$ is $\lambda$-presentable we have that any arrow $f: A \to C$ factors essentially uniquely through the diagram. We very often only use the factorisation, but as is clear in the answer above: sometimes the essential uniqueness is important. This means that if we have two different factorisations of $f$, so $f = d_i g = d_i g'$ for some $g, g': A \to D_i$, that then there must be some $j \in I$ with $i \leq j$ such that $d_{i,j} g = d_{i,j} g'$, where $d_{i,j}: D_i \to D_j$ is the arrow corresponding to $i \leq j$ in the diagram.
