How to show that any computable function is bounded above by Busy Beaver Function?
I think a sketch of the proof could be as follows.

First prove BB(k) is uncomputable. Assume it is computable, then it means there is some machine with some number of states m that can compute this value. Show that there exist a larger machine with some number of states M that calls this function on BB(M) and outputs BB(M)+1 1s, and then halts. Therefore there is a contradiction since it outputed more 1s than predicted by BB(M). It's easy to see M = m + C for some constant.

Now regarding the growth.

Assume G(k+1) >= BB(k+1) - BB(k) is computable for k >= K0.

Now, if the above is true then there must exist a machine with m states that computes G(k).

And there must also exist a machine, with a larger number of states M, that guesses BB(K0) and then computes UBB(M) = sum(K0..M)G(k) + BB(k0) and output UBB(M)+1 1s.

So this machine outputs more 1s than our assumed computable upper bound.

Therefore G(k) can not be computed for any K0.