Like so many problems in combinatorics, the key to this problem is understanding precisely what kind of things you are counting. You confusion seems to be arising, at least in part, from the fact that you are trying to count distributions of jobs to urns, and your authors are counting something seemingly different and abstruse to you. I will try to explain why they are in fact the same thing.
Consider the task of distributing 4 jobs among 26 urns as posed in your problem. If you were to naively count distributions of jobs to urns, you might start building distributions as follows in terms of sequences, where the value at the $i$th index in this list represents the number of jobs distributed to the $i$th urn:
$(1, 1, 1, 1, 0, 0, \ldots, 0)$
$(1, 1, 1, 0, 1, 0, \ldots, 0)$
$(1, 1, 1, 0, 0, 1, \ldots, 0)$
...
For a problem this large, you would get frustrated pretty quickly. While it's a nice idea, this type of solution doesn't seem to get us anywhere. So we have to conceptualize the problem in a different way, which leads me to your authors suggestion. Instead of building sequences, a common technique in combinatorics, why not instead pictorally represent the problem with o's and |'s, where o's represent jobs, and |'s (or dividers) represent divisions into urns. That is to say, the first sequence posed would look something like:
o|o|o|o|||||||$\ldots$|
where the two parallel dashes with no 'o's in between would represent an urn with no jobs in it.
The reason this technique is powerful is because it allows us to now count the different kinds of symbolic sequences we can build in a way that represents exactly what we're looking for: distributions of jobs to urns. But how do we count these kinds of distributions?
Well, we can see that as there are 26 urns, and 4 jobs to distribute, there are going to be a total of 29 symbols in our sequence (not 30, because we only need 25 (i.e., 26- 1) dividers to represent division into 26 total urns). Since we require that four of these symbols are 'o's, i.e., jobs, then we see that all possible sequences are formed by "choosing" four of the symbols in the sequence to be 'o's, and allocating the rest to be dividers. These build all the possible distributions we are looking for, and for this particular problem, we see that the number of possible distributions is thus ${26 - 1 + 4 \choose 4}$.
You might rightfully protest, "Why are we choosing 'o's? Shouldn't this method work if we instead chose 25 of the symbols to be dividers?" Well, that's a perfectly reasonable objection; but by the symmetry of $n C r$, we in fact have that ${29 \choose 4} = {29 \choose 25}$. You can see that this type of thinking quickly generalizes to distributions of $n$ indistinguishable things to $k$ distinguishable people; there will be $n$ objects, $k-1$ dividers, and we choose $n$ of the symbols to be objects, so we get ${n + k - 1 \choose n}$ possible distributions.
This explanation might seem a bit overkill (it probably is!), but this type of thinking can help you solve a lot of distribution problems of this kind. The analogue in the case of non-negative vector solutions to $x_{1} + \cdots + x_{k} = n$ is in realizing that this is the same as distributing $n$ indistinguishable things among $k$ distinguishable people, where $x_{i}$ simply represents the number of "things" distributed to the $i$th person.
I hope this helps!
Edit: in response to your next question, I'm not sure why your authors would present it this way (typically, the flow of logic is reversed!) but I can certainly explain.
The key is in noting that strictly positive solutions require $y_{i} > 0$ for all $i$, i.e., $y_{i} \geq 1$. One way to ensure this is to simply add one to each element of the non-negative solutions; that is, set each $y_{i}$ equal to $x_{i} + 1$. Since each $x_{i} \geq 0$, we then have that each $y_{i} \geq 1$ as desired. However, if we wish to leave the number of distributions unchanged, we cannot change what we are doing combinatorially; hence, for each $y_{i}$, if we set $y_{i} = x_{i} + 1$, then we must add 1 to the right hand side of the expression $x_{1} + \cdots + x_{r} = n$ for each $i \in \{1, 2, \cdots, r\}$. Then we have $y_{1} + \cdots + y_{r} = n + 1 + \cdots + 1 = n + 1*r = n + r$.
As a sort of combinatorial exercise, one way you can convince yourself that the number of distributions is unchanged is to think of it as follows. Imagine instead of distributing $n$ objects to $r$ people, accepting distributions where some people get zero objects, we instead consider distributing one object to each of the $r$ people a priori in order to ensure that each person gets at least one object. But to leave the distribution unchanged, that would mean we would distribute $n+r$ objects, allocating one object to each of the $r$ people from the start - hence why in "equation form" we would set $y_{i} = x_{i} + 1$.
Hope this helps! Feel free to post if you need further clarification.
Best Answer
For concreteness, let us work with the specific numbers $8$ and $3$ mentioned in the post, though the argument is general.
We have $8$ identical candies, and we want to distribute them among $3$ kids, with some kid(s) possibly getting no candy. Call this Task A.
Task B goes as follows. Distribute $8+3$ candies among the kids, with each kid getting at least $1$ candy. Then take away a candy from each kid.
It is clear that there are just as many ways to carry out Task B as there are to carry out Task A. And by the analysis of Proposition 1, there are $\binom{8+3-1}{3-1}$ ways to carry out Task B.
Another way: Imagine $8+3-1$ slots in a row, like this: $$\square\quad\square\quad\square\quad\square\quad\square\quad\square\quad\square\quad\square\quad\square\quad\square$$ We will put place $8$ candies ("stars") in these slots, with the other $2$ slots serving as separators ("bars"), possibly adjacent. The number of ways of placing the candies is $\binom{8+3-1}{8}$. It is the same as the number of ways of choosing the $2$ slots to be left blank.
So the number of solutions is $\binom{10}{8}$, or equivalently $\binom{10}{2}$.
In general, the same analysis shows that the number of ways to distribute $n$ candies among $r$ kids is $\binom{n+r-1}{n}$, or equivalently $\binom{n+r-1}{r-1}$.