If two coins are flipped and one gets head, what is the probability that both get head?

I have a doubt because I think that once the result of the first coin is obtained, just simply await the outcome of the second, which is completely independent of the previous one, and then we have a chance of $\frac12$ to get a head again.

But someone else tells me that as the possible events are:

Head - Head
Head - Tail
Tail - Head
Tail - Tail

then when we get a head we restrict ourselves to the first three cases, so the probability would be $\frac13$.

What is the right way?

I know there's a difference between saying "first came head" to say "one of the two came head", but if we have the first fact, aren't we supposed to know which one is that came head?

The problem here is that your phrase "one gets head" is not precise enough.

If it means "at least one of the coins comes up heads", then indeed there are 3 equally likely possibilities (HT,TH,HH) out of which exactly 1 has both coming up heads. This means that the desired probability is $\frac13$.

If it means "out of the two coins A,B that were flipped, A comes up heads", then B is equally likely to come up heads or tails, and so the desired probability is $\frac12$.

I think this is one of the cases where logic/mathematics totally goes overboard on a trivial problem, and you (as well as your friend) are overthinking it. But it's fun anyway, so...

There are at least 3 possible answers which are equally correct, depending on how pedantically one tries to twist the wording one way or the other. But for practical purposes it does not make a lot of sense because there is only exactly one solution that is correct (and immediately obvious) in each situation.

The first obvious interpretation is that you toss two coins into the air at the same time (which is not what you describe!). There are 4 possible outcomes. One of these outcomes has two heads, and one has no heads at all. You have set up the precondition that one coin gets head, which rules out the "no heads" outcome, leaving 3 possible outcomes. Only one of the three has two heads in it, thus: 1/3. This is a dependent probability. It is also an example of a Monty Hall Problem.

The second obvious interpretation is you toss one coin, and it comes up head. That's the precondition. You could just as well not have tossed the first coin at all. You now toss the second coin. Alternatively, you can toss the two coins together, but ignore all cases where the precondition that the first one gets head isn't fulfilled.
Assuming the second coin is not weighted or a trick coin with two heads or such, the chance is, of course, 1/2. From the point of view of the second coin, the first coin doesn't exist at all. This is a single (independent) probability.

The third obvious solution is zero. If two coins are flipped and exactly one coin gets head, the probability of both coins getting head is zero. This is a smart ass probability.

It's called conditional probability:

Let $A$ denote the event of getting exactly $2$ heads.

Let $B$ denote the event of getting at least $1$ head.

Then the probability that $A$ will occur given that $B$ has occurred is:

$$\frac{P(A\cap B)}{P(B)}=\frac{1/4}{3/4}=\frac13$$

EDIT:

Please note that in this specific case, $A\cap B=A$.

Therefore, one could simply calculate $\frac{P(A)}{P(B)}$.

However, this is not always the case.

For example, consider the following question: