There are a couple issues here.
To begin with, you're conflating languages (just sets of symbols) with theories (sets of sentences using the symbols in the language, in addition to the "purely logical" symbols - the latter being equality, parentheses, Booleans and quantifiers, and variables). For example, $\mathsf{ZFC}$ is a theory in the language $\{\in\}$.
You're also mixing up definability (which is relative to a structure) with provable behavior (which is relative to a theory). Recall that a set $X\subseteq\mathfrak{A}^n$ is definable in $\mathfrak{A}$ iff there is some formula $\varphi(x_1,...,x_n)$ in the language of $\mathfrak{A}$ such that $$X=\{(a_1,...,a_n):\mathfrak{A}\models\varphi(a_1,...,a_n)\}.$$ (See this old answer of mine for more about this.)
Definability, not provability, is the notion which is important here: provability is irrelevant to Tarski's undefinability theorem.
(OK, that's not strictly true, but it's true at the beginning.)
In general definability is much "broader" than provable behavior in any sense. For example, Godel's incompleteness theorem says (for example) that there is some sentence $\sigma$ such that $\mathsf{PA}$ can't prove or disprove "$\sigma$ is a theorem of $\mathsf{PA}$," but the set of $\mathsf{PA}$-theorems is definable over the standard model $\mathfrak{N}=(\mathbb{N};+,\cdot)$ of $\mathsf{PA}$.
I'm not sure what your proposed definition of definability is getting at. One key issue with your proposed definition is that definability only makes sense for subsets of (Cartesian powers of) the structure; it doesn't make sense to ask whether $X$ is definable in $\mathfrak{A}$ if $X\not\subseteq\mathfrak{A}^n$ for some $n$. Note that this indicates an abuse of terminology in the present situation: when we say "$Th(\mathfrak{N})$ is undefinable in $\mathfrak{N}$" what we really mean is "$\{\#(\varphi): \varphi\in Th(\mathfrak{N})\}$ is undefinable in $\mathfrak{N}$." To drive this point home, note that every countable set is in bijection with some definable subset of $\mathfrak{N}$, so that wouldn't give a useful notion.
OK, now let's turn to the statement of Tarski's undefinability theorem. The usual "concrete" version of Tarski is the following: "$\{\#(\varphi):\mathfrak{N}\models\varphi\}$ is undefinable in $\mathfrak{N}$," where $\mathfrak{N}$ is as mentioned above the standard model of arithmetic and "$\#$" is the usual Godel numbering map. Note that - as indicated earlier - theories and provability are not mentioned here at all.
The result above may feel a bit arbitrary; this is especially true since we often want to apply Tarski outside of the realm of $\mathfrak{N}$ specifically (e.g. to the whole set-theoretic universe). Rather than proving a new version of Tarski each time we need it, there is a more general form of the theorem which says roughly that if $\mathfrak{A}$ is a structure and $f$ is a map from the set of sentences in the language of $\mathfrak{A}$ to $\mathfrak{A}$ satisfying some basic properties, then $$\{f(\varphi):\mathfrak{A}\models\varphi\}$$ is not a definable subset of $\varphi$.
Now we're finally in a position to talk about how it's used. Often it appears not literally as a step in a proof, but rather as an indicator that a given proof attempt will not work (and indeed that the opposite result is likely true). For example, the natural way to attempt to show that the function $F$ sending $n$ to the least natural number not definable in $\mathfrak{N}$ by a first-order formula of length $<n$ is definable in $\mathfrak{N}$ would be to try to directly express "the least natural number not definable in $\mathfrak{N}$ by a first-order formula of length $<n$" in the language of first-order arithmetic; but when unfolded, that relies on being able to talk about truth in a definable way, which Tarski says we can't do.
Note that this isn't yet a proof that $F$ isn't definable in $\mathfrak{N}$ - maybe there's some clever other way to define it - but it kills the obvious approach and strongly suggests that $F$ isn't definable. The actual proof that $F$ is undefinable however is strongly indicated by the proof of Tarski's theorem, so we generally get away with saying "By Tarski, $F$ isn't definable."
That said, Tarski also gets used more "directly." For an example I personally find quite neat see here, where a slight generalization of the usual "concrete" phrasing of the theorem can be used to prove the first incompleteness theorem.
Also, note that the way you've stated $(1)$ is incorrect: the least natural number $k_{90}$ not definable in $\mathfrak{N}$ by a formula with $<90$ symbols is definable in $\mathfrak{N}$, since indeed every natural number is definable in $\mathfrak{N}$, but the formula "the least natural number not definable in $\mathfrak{N}$ by a formula with $<z$ symbols" is not definable in $\mathfrak{N}$.
