My question here is actually about whether different degress of impredicativity matter? To show that, lets confine ourselves with the following predicative formalism.
Language: multi-sorted first order logic
Primitives: $0, S, =, \in$
Syntatical restrictions include: $x_i \in y_{i+1}; x_i=y_i, S(x_0) $.
For convenience if a variable is written in un-indexed manner, then it's of sort zero; so $x$ is in reality $x_0$. The symbol $\mathbb N$ shall be left unindexed, though it is an object of sort 1.
Axioms:
Extensionality: $\forall x_{i+1} \, \forall y_{i+1} \, (\forall z_i (z_i \in x_{i+1} \iff z_i \in y_{i+1}) \to \\x_{i+1}=y_{i+1})$
Comprehension: $\exists x_{i+1} \forall y_i (y_i \in x_{i+1} \iff \phi(y_i))$
Define: $\mathbb N= \{x: x=x\}$
Naturals: $0 \in \mathbb N \\ x \in \mathbb N \to S(x) \in \mathbb N \\ S(x)\neq 0 \\ S(x)=S(y) \to x=y $
Induction: $\forall X_1: 0 \in X_1 \land \forall n (n \in X_1 \to n+1 \in X_1) \to \mathbb N \subseteq X_1 $
Now $\phi(y_i)$ is predicative ($0$-impredicative) if and only if the maximal sort a variable occurring in it can take is $i$.
$\phi(y_i)$ is $n$-impredicative if and only if the maximal sort a variable occurring in it can take is $i+n$.
Call the the above theory with the restriction of comprehension to $n$-impredicative formulas as: $\sf TTP_n$
Now we know from $\sf NF$ that $\sf NFP$ is weaker than $\sf PA$, while $\sf NFI$ is as strong as second order arithmetic, while if we allow $2$-impredicativity then we get full $\sf NF$, and further increase of degree of impredicativity is idle, so we still get just $\sf NF$. [see here and here]
Is it the same situation with the above type theory? In other words, is $\sf TTP_2$ equal in strenght to $\sf TTP_{n+2} $ ?
Yes. Predicative type theory plus the axiom of union, which looks up only one type, is full impredicative type theory. The situation in type theory is the same as in NF.