Why does the set $\{1,3,5,7... ; 2,4,6,8...\}$ qualify as well-ordered? How to explain this notation?

Sets have no "inherent order". They can have a natural order (such as $\Bbb N$ or $\Bbb R$ having a natural ordering that we consider somehow "part of the set"), but they do not have an inherent ordering.

Writing $\{1,3,5,\dots;0,2,4,\dots\}$ is a terrible abuse of notation.

But it is indeed a well-ordering of $\Bbb N$ which is not its natural ordering (which is also a well-ordering). Here is an explicit definition:

$$m\prec n\iff \begin{cases}m\text{ is odd and }n\text{ is even}, &\text{ or}\\m\equiv n\pmod 2\text{ and }m<n.\end{cases}$$

To see it is a well-ordering, note that each of the subsets, evens and odds, is ordered in the natural way which makes that part a well-order, and so given a non-empty set, it either has a smallest odd number or it is a subset of the even numbers.

As suggested, a well-ordered set is one in which each subset has a first element. The natural numbers $\mathbb{N}$ with their usual relation $\le$ form a prototypical example:

$$0 < 1 < 2 < 3 < 4 < 5 < \cdots$$

however what I believe your example, written as

$$\{ 1, 3, 5, 7, \cdots ; 2, 4, 6, 8, \cdots \}$$

is supposed to indicate (what book or other reference is this from?) is a well-ordered set in which the order looks like this:

$$1 < 3 < 5 < 7 < \cdots < 2 < 4 < 6 < 8 < \cdots$$

in other words, it's all the odd numbers ordered in the usual way, and then after ALL of them, all the even numbers, again, ordered in the usual way. The trick is that every odd number is "less" than every even number in this new ordering, and thus shows a well-order which is different from the usual ordering of the natural numbers. It's meant to show that there are more forms a well-order can take than one might at first expect. In particular, it is a well-order which has an "internal infinity" in that there are an infinite number of elements between any odd number and any even number.

With regard to the question of the initial element of the subset $\{7, 2\}$, the answer is simple: looking at the sequence as shown above and where 7 and 2 fall therein, it is seen at a glance to be 7 (not 2). In terms of the mathematical definition, since 7 is odd, and 2 is even, 2 must be "greater" than every odd number, hence also greater than 7 and 7 is the initial element.