Determining a program's execution time by its length in bits?

This is a question popped into my mind while reading the halting problem, collatz conjecture and Kolmogorov complexity. I have tried to search for something similar but I was unable to find a particular topic maybe because it is not of great value or it could just be a trivial question.

For the sake of simplicity I will give three examples of programs/functions.

function one(s):
  return s

function two(s):
  while (True):
    print s

function three(s):
  for i from 0 to 10^10:
    print(s)

So my questions is, if there is a way to formalize the length of a program (like the bits used to describe it) and also the internal memory used by the program, to determine the minimum/maximum number of time/steps needed to decide whether the program will terminate or run forever.

For example, in the first function the program doesn't alter its internal memory and halts after some time steps.
In the second example, the program runs forever but the program also doesn't alter its internal memory. For example, if we considered all the programs with the same length as with the program two that do not alter their state, couldn't we determine an upper bound of steps, which if surpassed we could conclude that this program will never terminate ? (If not why ?)
On the last example, the program alters its state (variable i). So, at each step the upper bound may change.

[In short]
Kolmogorov complexity suggests a way of finding the (descriptive) complexity of an object such as a piece of text. I would like to know, given a formal way of describing the memory-space used by a program (computed in runtime), if we could compute a maximum number of steps, which if surpassed would allow us to know whether this program will terminate or run forever.

Finally, I would like to suggest me any source that I might find useful and help me figure out what I am exactly looking for.
Thank you. (sorry for my English, not my native language. I hope I was clear)

If a deterministic Turing machine enters precisely the same configuration twice (which we can detect b keeping a trace of configurations seen so far), then we immediately know the TM will loop forever.

If it known in advance that a deterministic Turing machine cannot possibly use more than some fixed constant amount of its input tape, then the TM must explicitly halt or eventually enter some configuration it has already visited. Suppose the TM can use at most k tape cells, the tape alphabet is T and the set of states is Q. Then there are (|T|+1)^k * |Q| unique configurations (the number of strings over (T union blank) of length k times the number of states) and by the pigeonhole principle we know that a TM that takes that many steps must enter some configuration it has already been to before.

one: because we are given that this function does not use internal memory, we know that it either halts or loops forever.

two: because we are given that this function does not use internal memory, we know that it either halts or loops forever.

three: because we are given that this function only uses a fixed amount of internal memory (like 34 bits) we can tell in fewer than 2^34 iterations of the loop whether the TM will halt or not for any given input s, guaranteed.

Now, knowing how much tape a TM is going to use, or how much memory a program is going to use, is not a problem a TM can solve. But if you have an oracle (like a person who was able to do a proof) that tells you a correct fixed upper bound on memory, then the halting problem is solvable.