Let's examine what the book says: To show that $\mathbb N\times\mathbb N$ is countable, we need to exhibit a bijection from $\mathbb N\times \mathbb N$ onto $\mathbb N$.
The authors of the book describe the bijection by saying: Enumerate the pairs $(m,n)$ according to the increasing sum $m+n$ and, for each fixed $m+n$, according to increasing $m$.
[Just for fun, let's list some of the pairs in order: Since $m,n\in\mathbb N$, then $m+n\ge2$, so the first value of $m+n$ is $2$. There is only one pair $(m,n)$ with that sum, namely $(1,1)$. The next possible value for $m+n$ is $3$. The equation $m+n=3$ has two solutions in the natural numbers, namely $m=2,n=1$ and $m=1,n=2$. The description tells us we first list $(1,2)$ and then $(2,1)$. The next value is $m+n=4$, and the pairs of solutions are listed as $(1,3),(2,2),(3,1)$. They will be followed by $(1,4),(2,3),(3,2),(4,1)$, in that order. Etc. So the list begins: $$(1,1),(1,2),(2,1),(1,3),(2,2),(3,1),(1,4),(2,3),(3,2),(4,1),(1,5),\dots$$ For example, this means that if $f$ is the function we have described, then $f(2,1)=3$, and if $f(a,b)=10$, then $(a,b)=(4,1)$.]
Does this assign a natural number as a value to each pair $(m,n)$ of natural numbers? Yes.
(In fact, there is an easy algorithm that allows us to find that value, although that is irrelevant: The algorithm itself is not very imaginative: Just go on listing first the pairs with sum $2$, then those with sum $3$, then those with sum $4$, and so on, until $(m,n)$ is listed, and then count the elements that have been listed so far to find where in the list $(m,n)$ appears, and that is the number assigned to it.)
Is this assignment unambiguous? Yes.
Of course. The description is unambiguous, there is no leeway at all in how to produce the list. So, so far, we know that we have a function with domain all of $\mathbb N\times\mathbb N$ and range in $\mathbb N$.
[We can say more if needed, but I think this is just clutter at some point: For each fixed natural number $k$, there are only finitely many pairs $(m,n)$ with $m+n=k$. In fact, we can identify them all, since $m+n=k$ implies that $m\le k$, so $m$ is one of $1,2,\dots,k-1,k$, and for each of these options there is a unique value of $n$ that works, namely $n=k-m$. What does that buy us? Well, given any pair $(m,n)$, if we define $k$ by $k:=m+n$, there are only finitely many pairs $(a,b)$ with $a+b\le k$, so we can list them in a finite list, in the ordering described -- first order by the size of $a+b$, and among those with the same $a+b$, order by the size of $a$ -- so that indeed we assigned to $(m,n)$ a number, because finite unions of finite sets are finite (or, more to the point in our case, the concatenation of finitely many finite lists is still a finite list), and this number is well defined. If pressured, we could produce a formula for what number it is that we assigned to $(m,n)$ but, again, there is no need.]
If a natural number $n$ is assigned to a pair $(a,b)$, is this the only pair that $n$ is assigned to? Obviously.
Any list with at least $n$ elements has only one element in the $n$th position. This means that the function we have is injective.
Is the range of this function an initial segment of $\mathbb N$? Yes.
We are listing the numbers, after all. This means we are not skipping any values. If $f(a,b)=20$, then the list has assigned the numbers $1,2,\dots,19$ to other pairs $(m,n)$ as well.
Is the range all of $\mathbb N$? Yes.
Obvious: For any $k$, according to the description of the list, thee pairs $(1,1),(2,2),\dots,(k,k)$ are listed in that order, with possibly many other numbers interpolated between them. So the list of numbers has length at least $k$, so some pair has been assigned the number $k$. We have now shown that the function we have is also surjective.
That's it, that's the proof. Can it be expanded? Of course. We can add much to it, but it seems utterly unnecessary. My point here is that, as explanations go, there is really not much to say. The bijection was indeed unambiguously and precisely defined (in spite of what the authors say).
The sketch in the book is not a full proof: It did not argue that this description indeed gives us a function, that its domain is all of $\mathbb N\times\mathbb N$, that the function is injective, its range is an initial segment of $\mathbb N$ and, in fact, that this initial segment is all of $\mathbb N$. Those are all the details we needed to verify to check we had a bijection. But, as these things go, we really were not missing much in terms of details anyway.
A piece of advice: Read the details of the proof in the book, even if "long and a bit scary". If nothing else, out of curiosity. If the book is not saying more than what I say above, then at the end we see that there was not much to be scared of. If the book goes into additional details, at the end we may know even more about this bijection, and understand it better. Perhaps the book will expand in great detail some of the items I essentially called trivial in this write-up, and perhaps in the context of the book, it makes sense that these details are needed, and not quite as trivial as I believe. (But if this is the case, it will only make sense if you are familiar with the book, and with what the authors allow themselves to assume through it. It will not be something you can discern just from reading the details in that appendix.)
Or perhaps you will see that the authors are worrying unnecessarily, which may well be the case. Whichever the outcome, it seems you'll gain from the experience. As you can see from the comments and the many answers, it is not that there is universal consensus among mathematicians on what we allow ourselves to assume (even depending on the audience). That is always interesting to experience, I think it makes mathematics feel more real and alive, which is always good to realize.