You can't assume "the boy is k1". The original 2 * 7 * 2 * 7 cases were indeed all equiprobable different cases. And we're not given that K1 is a boy. We're given that at least one child is a boy.

If K1 is a girl and K2 is a boy born on Tuesday, this still counts as the family (Mr. Jones, if you like) having a boy born on Tuesday. There are 27 cases that count as the family having a boy born on Tuesday, all equiprobable. And out of those, 14 also count as the family having a girl.

As for your noting that we can split these cases even more finely, so that there's no distinguished end-all, be-all partitioning of cases, sure, you can do that. What I'm really saying is this:

1/2 of two-child families have their elder child being a boy. 1/2 of two-child families have their elder child being a girl. [On conventional idealizations for these problems. You surely do not dispute this, yes? You may not care about this number, but you don't dispute it, right?]

In each of those subgroups, 1/7 of families have their elder child born on Sunday, 1/7 have their elder child born on Monday, etc. [Do you dispute this?]

In each of THOSE subgroups, 1/2 of families have their younger child a boy, and 1/2 have their younger child a girl. [Any dispute?]

And in each of THOSE subgroups, 1/7 of families have their younger child born on Sunday, 1/7 of families have their younger child born on Monday, etc. [Any dispute?]

And some amount of those have low birth weight, some have high birthweight, some have 5 freckles, etc., but we needn't figure out those numbers.

So now I've carved the world up into 2 * 7 * 2 * 7 groups, based on gender and birth-date-of-week for older and younger child. We can carve the world up into groups in different ways also, more finely or more coarsely or just differently. But making the four conventional assumptions we just made, the 2 * 7 * 2 * 7 grouping based on gender and birth-date-of-week for older and younger child is such that each particular such group takes up 1/2 * 1/7 * 1/2 * 1/7 of all families; these are all equifrequent groups.

And that having been done, we find that in 27 of these groups, there is at least one boy born on a Tuesday. In 13, the elder child is a boy born on Tuesday but not the younger child; in 13, the younger child is a boy born on Tuesday but not the elder child; in 1, both children are boys born on Tuesday.

But the question was not intended to be about a specific boy. The question was intended to be "Out of families that have a boy born on Tuesday (meaning at least one boy born on Tuesday), what proportion have a girl?". Any family with at least one boy born on Tuesday counts as having "a boy born on Tuesday", and even families with two boys born on Tuesday count, with no particular of their two boys given any distinguished status.

Perhaps you read the question differently; that, then, is a problem with the phrasing of the question for communicating to you its intent. But when it was asked "What is the probability Mr. Jones has a girl, given that he has a boy born on Tuesday?", what the author indeed intended this to mean, and would be generally taken in the conventional language of probability to mean, was "Out of families that have at least one boy born on Tuesday, what proportion have a girl?".

And we find that, out of the 27 equally sized groups of families that have at least one boy born on Tuesday, 14 of them have a girl, so that the answer to this question becomes 14/27.