After expressing the desired answer as a ratio of unconditional probabilities:
$$P(\text{two girls}\mid\text{at least one winter girl})
=\frac{P(\text{two girls}\mathop{\textit{AND}}\:\text{at least one winter girl})}{P(\text{at least one winter girl})}$$
we need to actually calculate those unconditional probabilities to get the answer.
The bottom probability is a pretty standard calculation: the probability of at least one winter girl is one minus the probability of no winter girls, and "no winter girls" is a pair of independent "not a winter girl" events, so that's how we get $1-(7/8)^2 = 15/64$. But you knew that -- it's the standard way of calculating an "at least once" probability, and the probability that you've seen it used at least once is close to one (ha ha).
The top probability is harder. One way to calculate it directly would be to enumerate all possible outcomes of 4 seasons and 2 sexes for the first child, and another 4 seasons and 2 sexes for the second child, so $4 \times 2 \times 4 \times 2 = 64$ outcomes, which we assume are equally probable. Then, we can exhaustively list the outcomes that match the event described in the top probability. If we write WXYZ for the seasons (W for winter) and MF for the sexes, the subset of 64 outcomes matching this event are the seven outcomes:
WGWG, XGWG, YGWG, ZGWG, WGXG, WGYG, WGZY
Every other outcome either includes a boy (so not two girls), or is one of the nine outcomes with two girls but no winter girls (e.g., XGXG, XGYG, etc.) So, the top probability is 7/64, which gives the correct final answer to the problem of $7/15$.
But, that's really tedious!
If we want to calculate the top probability less tediously, we have to reach into our toolbox for calculating probabilities of boolean combinations of events. We don't have a lot of tools for this. We always have $P(\mathop{\mathit{NOT}} A) = 1-P(A)$. For $\mathit{OR}$ing events, we have $P(A\cup B)=P(A)+P(B)$ if the events are mutually exclusive and $P(A\cup B) = P(A) + P(B) - P(A\cap B)$ if the events are independent. For $\mathit{AND}$ing events, I guess we have $P(A\cap B)=0$ if the events are mutually exclusive and $P(A\cap B) = A$ if $A\subseteq B$, but the only really useful one is $P(A\cap B)=P(A)P(B)$ if the events are independent.
This is what motivates trying to construct an equivalent event that consists of an intersection of independent events. The approach taken in the textbook solution is a reasonably natural way of constructing an equivalent event that is structurally similar to the original event but involves an intersection of independent events. If you look at the two sub-events in the original event, the obvious problem is that "girl" appears in both the left sub-event "two girls" and the right sub-event "at least one winter girl", and this introduces a dependency. If we want to remove this dependency, the natural thing to do is to remove "girl" from one of the events. Removing it from "two girls" won't work, but it turns out that we can remove it from "at least one winter girl" without actually modifying the event.
Other approaches work, too. We are just trying to rewrite the top probability into something that matches our "toolbox" so we can evaluate it.
For example, we could observe that the top event is the union of the mutually exclusive events "both girls are winter girls", "first is a winter girl, second is a non-winter girl", and "first is a non-winter girl, second is a winter girl". By independence of the two girls, these have probabilities $1/8 \times 1/8=1/64$, $1/8 \times 3/8 = 3/64$ and $3/8 \times 1/8 = 3/64$ respectively, and by the mutual exclusivity of the events, the probability of their union is the sum: $1/64 + 3/64 + 3/64 = 7/64$.
Or, we could observe that the events "two girls AND at least one winter girl" and "two non-winter girls" are mutually exclusive and have union equal to the event "two girls". This means:
$$P(\text{two girls}) = P(\text{two girls} \mathop{\mathit{AND}} \text{at least one winter girl}) + P(\text{two non-winter girls})$$
Since $P(\text{two girls})=(1/2)^2=16/64$ and $P(\text{two non-winter girls}) = (3/8)^2=9/64$, we can solve for the desired probability to get $16/64-9/64=7/64$.
Or, we could observe that top probability is the union of the mutually exclusive events "first child is a winter girl and second child is a girl" and the "first child is a non-winter girl and second child is a winter girl". These have probabilities $1/8 \times 1/2 = 1/16 = 4/64$ and $3/8 \times 1/8 = 3/64$ respectively, and sum them to again get $7/64$.
Hopefully you can see that none of these truly "come out of nowhere". There's even a reasonable chance that the textbook authors actually tried a few of these methods before eventually choosing the most "elegant" one to include as their solution.
