I really don't know exactly what you are asking, but this is too long for a comment.
Looking at the mapping $\Psi$ on p.162 instead as a function $\mathcal{P} (F) \to \mathcal{P} (F)$, it is not too difficult to see that $\Psi (A) = \{ \phi \in A : \phi \text{ has free variables } v_0, \ldots , v_{n-1} \}.$ That is, $\Psi(A)$ is just the subset of $A$ consisting of those formulae with the "correct" free variables.
If $D$ is an $F$-diagram, then there is a countable model $\mathcal{M}$ (with universe $\omega$) such that $D = \{ \phi (v_0, \ldots, v_m) \in F : \mathcal{M} \models \phi (0,1,\ldots,m) \}.^{\text{(*)}}$ Then $\Psi (D)$ would consist of those $\phi \in D$ whose free varaibles are among $v_0, \ldots , v_{n-1}$. Adjusting some indices in p.159, it is clear that $\Psi (D)$ is an $F$-type: it is the set of all formulae satisfied in $\mathcal{M}$, above, by the $n$-tuple $(0,\ldots,n-1)$.
$^{\text{(*)}}$ I take it that formulae in an $F$-diagram can have free variables any (finite) initial segment of the variable symbols, which seems to better match common usage. Marker's presentation seems to imply that the formulae in an $F$-diagram have free variables only among $v_0, \ldots , v_n$, where $n$ is the same natural number as in the definition of the set $S_n(F,T)$ we are concerned with.