The two-daughter-problem
I think the confusion arises because the classical boy-girl problem is ambiguous:
'You know that Mr.Smith has two kids, one of which is a girl. What is the chance she has a sister?'
The ambiguity here is that from this description, it is not clear how we came to know that 'Mr.Smith has two kids, one of which is a daughter.'
Consider the following two scenarios:
Scenario 1:
You have never met Mr. Smith before, but one day you run into him in the store. He has a little girl with him, which he tells you is one of his two children.
Scenario 2:
You are a TV producer, and you decide to do a show on 'what is it like to raise a daughter?' and you put out a call for such parents to come on the show. Mr.Smith agrees to come on the show, and as you get talking he tells you that he has two children.
Now notice: the original description applies to both cases. That is, in both cases it is true that you know that 'Mr.Smith has two children, one of which is a daughter'.
However, in scenario 1, the chance of Mr. Smith having two daughters is $\frac{1}{2}$, but in scenario 2 it is $\frac{1}{3}$. The difference is that in the first scenario one specific child has been identified as female (and thus the chance of having two daughters amounts to her sibling being female, which is $\frac{1}{2}$), while in the second scenario no specific child is identified, so we can't talk about 'her sibling' anymore, and instead have to consider a conditional probability which turns out to be $\frac{1}{3}$.
Now, your original scenario, where you don't know anything about Mr. Smith other than that he has two children, and then Mr.Smith says 'I am so happy Victoria got a scholarship!' is like scenario 1, not scenario 2. That is, unless Mr. smith has two daughters called Victoria (which is possible, but extremely unlikely, and if he did one would have expected him to say something like 'my older Victoria'), with his statement Mr.Smith has singled out 1 of his two children, making it equivalent to scenario 1.
Indeed, I would bet that most real life cases where at some point it is true that 'you know of some parent to have two children, one of which is a girl' are logically isomorph to scenario 1, not scenario 2. That is, the classic two-girl problem is fun and all, but most of the time the description of the problem is ambiguous from the start, and if you are careful to phrase it in a way so that the answer is $\frac{1}{3}$, you will realize how uncommon it is for that kind of scenario to occur in real life. (Indeed, notice how I had to work pretty hard to come up with a real life scenario that is at least somewhat plausible).
Finally, all the variations of whether Victoria is the oldest, youngest, or whether you don't even know her name ('Mr. Smith tells you one his children got a scholarship to the All Girls Academy') do not change any of the probabilities (as you argued correctly): in most real life scenarios, the way you come to know that 'Mr.Smith has two children, one of which is a girl' (and I would say that includes your original scenario) means that the chance of the other child being a girl is $\frac{1}{2}$, not $\frac{1}{3}$.
So, when at the end of you original post you ask "where is my error?" I would reply: your 'error' is that you assumed that the correct answer should be $\frac{1}{3}$, and that since your argument implied that is would be $\frac{1}{2}$, you concluded that there must have been an error in your reasoning. But, as it turns out, there wasn't! For your scenario, the answer is indeed $\frac{1}{2}$, and not $\frac{1}{3}$. So your 'error' was to think that you had made an error!
Put a different way: you were temporarily blinded by the pure math ( and I say 'temporarily', because you ended up asking all the right citical questions, and later realized that the classic two-girl problem is ambiguous: good job!). But what I mean is: we have seen this two-girl problem so often, and we have been told that the solution is $\frac{1}{3}$ so many times, that you immediately assume that also in your descibed scenario that is the correct answer... When in fact that is not case because the initial assumptions are different: the classic problem assumes a Type 2 scenario, but the original scenario described in your post is a Type 1 scenario.
It's just like the Monty Hall problem ... We have seen it so often, that as soon as it 'smells' like the Monty Hall problem, we say 'switch!' ... when in fact there are all kinds of subtle variants in which switching is not any better, and sometimes even worse!
Also take a look at the Monkey Business Illusion: we have see that video of the gorilla appearing in the middle of people passing a basketball so many times that we can now surprise people on the basis of that!
Let us take a pragmatic approach to this. For the first problem:
Step 1: Round up a million men, each of whom has two children.
Step 2: Tell all of the men who have no daughters to go home.
Step 3: Ask all of the remaining men who have two daughters to raise their hands.
Obviously, about one third of the remaining men will raise their hands: about 750,000 men remain, and about 250,000 of them have two daughters.
For the second problem:
Step 1: Round up a million men, each of whom has two children.
Step 2: Tell all of the men who don't have a daughter named Victoria to go home. (We can ignore the scholarship.)
Step 3: Ask all of the remaining men who have two daughters to raise their hands.
Now, suppose 1 in 100 girls are named Victoria. (The exact figure doesn't matter.) Then of the 500,000 fathers with a daughter and a son, 5,000 of them will have daughters named Victoria; and of the 250,000 fathers with two daughters, 5,000 of them will also have a daughter named Victoria (because they have 500,000 daughters in total). Therefore, of the 10,000 men remaining, 5,000 will raise their hands.
So the probability that Mr Smith has two daughters is $1/2$.
Let $XY$ denote that the sex of the younger sibling is $X$ and that of the older sibling is $Y$. $X$ and $Y$ may be $M$ or $F$, male and female. We have the following three equally likely elementary events
$$\{FF, FM, MF, MM\}.$$
These are equally likely, so $P(\{XY\})=\frac14$ for all possible $X,Y$.
The event that at least one of the siblings is a girl is
$$\{FF, FM, MF\}.$$
The event that both siblings are females is
$$\{FF\}.$$
We want to calculate the following conditional probability
$$P(\{FF\}\mid \{FF, FM, MF\})=\frac{P(\{FF\})}{P(\{FF, FM, MF\})}=\frac{\frac14}{\frac34}=\frac13.$$
The question remains: Do we agree that the following two questions are the same questions?
What is the probability that in a family both children are girls assuming that at least one of the children is a girl?
Assume that in a family of two children one of the children is a girl. What is the probability then that the other child is also a girl?
EDIT
Assume that a father says that he has a daughter and that daughter is older than the other child of his. Then our question modifies:
Assume that the older kid is a girl, what is the probability that the younger child is also a girl. Our conditional probability is then:
$$P(\{FF\}\mid \{FF, MF\})=\frac{P(\{FF\})}{P(\{FF,MF\})}=\frac{\frac14}{\frac12}=\frac12.$$
So, there is no contradiction. The second question is simply another question.
EDIT 2
I am only thinking... I realize that whatever the most honorable father's answer is the probability changes to $\frac12$. Wrong! Let's see what if I don't get an answer. Then the answer is either yes or no. That is, we have the following conditional probability:
$$P(\{FF\}\mid \{FF, MF\}\cup\{FF, FM\})=\frac{P(\{FF\})}{P(\{FF, FM, MF\})}=\frac{\frac14}{\frac34}=\frac13.$$