Galois types, factorization I do not follow here on the page $13$ in the
proof of proposition $6.2$
what does it mean that $f$ factors over $f_0$ , why such $f_0$ exists and how it relates to definitions $4.1$ and $6.1$ Also why does such $N_0$ $\lambda$-presentable
exist.


 A: That "$f$ factors over $f_0$" just means that there are $f_0: M \to N_0$ and $h: N_0 \to N$ such that $f = h f_0$. We also often write "$f$ factors through $f_0$" ("factoring over" seems to be more model-theoretic language).

Now for why such $N_0$ exists. We may assume our category is $\lambda$-accessible. Then $N$ is a $\lambda$-directed colimit of $\lambda$-presentable objects. That is
$$
N = \operatorname{colim}_{i \in I} N_i
$$
for some $\lambda$-directed diagram $(N_i)_{i \in I}$. Since $M$ is $\lambda$-presentable, the arrow $f: M \to N$ will factor as $M \xrightarrow{f_i} N_i \xrightarrow{n_i} N$ for some $i \in I$. This is not quite the $N_0$ from the proof yet, because we may not have $a \in U(N_i)$. Our category also has concrete directed colimtis. Since a $\lambda$-directed colimit is in particular a directed colimit, this means that
$$
U(N) = \bigcup_{i \in I} U(N_i).
$$
So there must be $j \in I$ such that $a \in U(N_j)$. Now let $\ell \in I$ be such that $i,j \leq \ell$. We can now take $N_0$ to be $N_\ell$ and $f_0$ to be the composition $M \xrightarrow{f_i} N_i \xrightarrow{i \leq \ell} N_\ell$. Note that the arrow $N_0 \to N$, which I called $h$ at the start of this answer, is just the coprojection $N_\ell \to N$.
This really is a category-theoretic version of the usual downward Löwenheim-Skolem argument. There we would have that $\lambda$ is the Löwenheim-Skolem number (or technically, the next cardinal). What this is then saying is that for $M \preceq N$ and $a \in N$ with $|M| < \lambda$ there is already $N_0 \preceq N$ with $|N_0| < \lambda$ such that $M \subseteq N_0$ (and hence $M \preceq N_0$) and $a \in N_0$.

I'll try to give a bit of intuition about the rest of the proof and how it connects with the definitions you asked about.
The relation with definition 4.1 is that by construction $(f_0, a)$ has the same Galois type, or "is equivalent" in the language of definition 4.1, as $(f, a)$. This is simply witnessed by the identity on $N$ and the arrow $N_0 \to N$ (which I called $h$ before). So basically we want to find a representative of the same Galois type, but in a 'small' (i.e. $\lambda$-presentable) model.
Then we can use that to prove that $K$ is indeed $\lambda$-Galois saturated (as in definition 6.1). Namely, since $N_0$ is small enough we can embed it entirely in $K$, while respecting the way $M$ is embedded in it already (this is what it means to be $\lambda$-saturated in this context, which $K$ is). That is, we already had $g: M \to K$ and now we get $g_0: N_0 \to K$ such that $g = g_0 f_0$, i.e. embedding $M$ directly in $K$ or via $N_0$ is the same. Finally, composing the pair $(f_0, a)$ with $g_0$ yields $(g_0 f_0, U(g_0)(a))$ and these pairs are equivalent (this is a general fact for these Galois types, follows directly from the definition). So by construction $(f, a)$ and $(f_0, a)$ are equivalent and $(f_0, a)$ and $(g, U(g_0)(a)) = (g_0 f_0, U(g_0)(a))$ are equivalent. This 'equivalent' really is an equivalence relation because we assume amalgamation (assumption 4.2) implies transitivity for this relation (reflexivity and symmetry are trivial), again a nice exercise. So in the language of definition 6.1 $(f, a)$ is realised in $K$, which is what had to be proved.
