You are always allowed to use transfinite induction in your proofs.
There is a problem, however, that sometimes the limit stages require the axiom of choice to hold.
The transfinite induction argument simply says something like induction, only general: suppose that you walked over all ordinals until $\alpha$, how do you plan on stepping through $\alpha$?
In the "usual" induction we usually just define the next step, or show that our idea is well-defined. It is commonly just as simple for successor ordinal steps in transfinite induction. However it is often the case that there is no well-defined solution for the limit ordinal. When something like this happens we must use the axiom of choice to help us show that there exists a well-defined solution for this problem.
Regardless to that, however, we may use transfinite induction whenever we would like. The question should have been phrased when can we use it, and the answer to that is whenever you are able to show a well-defined step for every ordinal. Sometimes it is constructive and other times it requires the axiom of choice. To such a general question there is no particularly pinpointed answer.
To the edit:
I think that you want to ask when in regular mathematics (i.e. not set theory) the induction follows to the limit case, that is "to the infinite step". The answer is, again, in its general form "when it can".
Sometimes it is possible and sometimes it is not possible. However distinguishing between those two cases requires a lot of training and usually a deep understanding of how the mathematical objects in question behave.
Again, if you wish to present a particular case then it can be explained further why and when it is possible to go beyond the finite steps; however just asking for a general explanation won't work.
Mathematics is not a bunch of algorithm and rules, it is a sea of definitions in which we swim constantly, and to wade between those definitions one has to have insights.