The key idea Feferman is exploiting is that there can be two different enumerations of the axioms of a theory, so that the theory does not prove that the two enumerations give the same theory.
Here is an example. Let $A$ be a set of the axioms defined by some formula $\phi(n)$ (that is, $\phi(x)$ holds for exactly those $x$ that are in $A$). Define a new formula $\psi(n)$ like so:
$\psi(n) \equiv \phi(n) \land \text{Con}( \langle x \leq n : \phi(x)\rangle)$
Where Con(σ) is a formula which says that no contradiction is provable from the axioms listed in the sequence σ.
In the case where $A$ is the set of axioms for a suitable consistent theory $T$ that satisfies the second incompleteness theorem, the following hold:
(1) In the standard model, we have $\phi(n) \Leftrightarrow \psi(n)$ for all $n$, because $T$ really is consistent.
(2) T does not prove that $\phi(n) \Leftrightarrow \psi(n)$ for all $n$, because this equivalence implies that T is consistent.
(3) If we use $\psi$ to define a formula Conψ(T), then T will prove (under the assumption that the empty theory is consistent, if this is not provable in T) that the theory defined by ψ is consistent. However, T will not prove Conφ(T), which is the usual consistency sentence for T.
This kind of example is presented in a more precise way in Feferman's 1960 paper that you mentioned, along with precise hypotheses on the theory and sharper results.
My opinion is that we cannot regard a proof of Conψ(T) as a proof of the consistency of T, because although φ and ψ are extensionally identical, they do not intensionally represent the same theory. Feferman expresses a similar idea on his p. 69. Of course, this is a matter of philosophy or interpretation rather than a formal mathematical question.
Addendum
The difference between extensional and intensional equality is easiest to explain by example. Let A be the empty set and let B be the set of integers larger than 1 that do not have a unique prime factorization. Then we know B is also the empty set: so A and B are extensionally equal. But the definition of B is much different than the definition of A: so A and B are intensionally different.
This distinction is often important in contexts like computability and formal logic where the things that you work with are actually definitions (also called codes, indices, Gödel numbers, or notations) rather than the objects being defined. In many cases, extensional equality is problematic, because of computability or effectiveness problems. For example, in my answer above, we know that φ and ψ define the same set in the real world, but this fact requires proof, and that proof may be impossible in the object theory we are dealing with. On the other hand, intensional equality is easy to decide, provided you are working directly with definitions.
Best Answer
Nice questions!
For the first question, I claim that every theory $T$ is $k$-incomplete in your sense for every finite $k$. This is because every statement appears explicitly as a part of any proof of it, and when $\phi$ is any very long theorem of $T$ in comparison with $k$, then any proof of $\phi$ has at least as many symbols in it as $\phi$ does, and so by your measures $\phi$ has no proof of size $k$ and hence is $k$-Goedelian by your definition. Thus, for any given $k$, the theory $T$ is $k$-incomplete. In particular, every theory becomes $k$-incomplete.
For the second question, it seems that any decent consistent theory will prove its own $k$-consistency for any particular $k$. For example, if $T$ is consistent and extends $PA$, and probably even extending $Q$ suffices, then since $T$ really does not have any proofs of contradictions, it has none of size $k$, and since there are only finitely many proofs of this size, which can be enumerated and known to be an exhaustive list within the theory $T$, it follows that $T$ will prove that there is no proof of a contradiction in $T$ of size at most $k$. That is, for each $k$ separately, $T$ proves its own $k$-consistency.
But since this is not a $k$-proof of its own consistency, it doesn't quite answer your second question. But it shows that you have to pay attention to the precise measure of $k$-provability. For example, it isn't even clear that a $k$-inconsistent theory will necessarily $k$-prove much, since perhaps the $k$-proof of a contradiction already uses up most of the $k$ symbols, leaving little room for further deductions from that contradiction.
In light of this, we can make a negative answer to your second question as follows. Let $T$ be the theory PA + $0=1$, which is $k$-inconsistent for a very small value of $k$, perhaps $25$ or so, since the extra axiom directly contradicts an axiom of PA. But this $k$ is much too small to even state the $k$-consistency of $T$, and so $T$ will not $k$-prove its own $k$-consistency. So it violates the second $k$-incompleteness theorem, because this is a theory that is $k$-inconsistent but does not $k$-prove its own $k$-consistency, contrary to the proposed equivalence in your question.