Assume you have some notion of proof complexity: for instance, at the basic level, the length of a proof, or the number of symbols used, take your pick (there are more involved measures, but for sake of simplicity I will limit myself to the above).


Now, start from some ground arithmetical theory $T$, and say that $T\vdash\_k \phi$ if there is a proof of $\phi$ from $T$ of complexity $\leq k$.


I call a sentence $\phi$ $k$-godelian iff it is $k$-undecidable (i.e. neither $T\vdash\_k \phi$ nor $T\vdash\_k \neg\phi$ , but provable in the standard sense in the theory $T$ (and so true in $N$, for those fortunate and seemingly numerous mortals who believe in such a creature).


$T$ is $k$-incomplete if there is such a $\phi$.


Now the two questions:


1. (first k-incompleteness) Which $T$s are k-incomplete for every $k$?


Are there theories that *become* eventually $k$-incomplete for a sufficiently large $k$ ?
2. (second k-incompleteness). Does any $k$-incomplete theory also satisfy the $k$-version of Godel 's second incompleteness? That is, is it true that $T$ is such that it $k$-proves its $k$-consistency only if it is $k$-inconsistent?



Finite versions of  Godel' s incompleteness
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.


