In a loose sense $S$ is the set of possible remainders: it’s the set of amounts by which $n$ differs from a multiple of $m$. That is, you look at all possible multiples $qm$ of $m$ (for $q\in\Bbb N$) and see how much $n$ differs from each of them; those differences are the members of $S$. As a result $S$ contains the non-negative numbers in the following list:
$\begin{align*} &n-0\cdot m=n\;,\\ &n-1\cdot m=n-m\;,\\ &n-2\cdot m=n-2m\;,\\ &n-3\cdot m=n-3m\;,\\ &\qquad\qquad\;\,\vdots \end{align*}\tag{1}$
Remember, we’re looking for natural numbers $q$ and $r$ such that $n=qm+r$, which means that $r=n-qm$; thus, if such $q$ and $r$ exist, the remainder $r$ must be one of the numbers in the list $(1)$. We also want $r$ to be non-negative, so $r$ will actually be in $S$ and not just in the list $(1)$.
We know that $n\in\Bbb N$, so we know that the list $(1)$ does have at least one non-negative member, and therefore $S\ne\varnothing$. You ask how $n$ can be a candidate for the remainder: what if $m>n$? What if, for instance, $m=5$ and $n=3$? Then dividing $n$ by $m$ yields a quotient $q=0$ and a remainder $r=n=3$: $3=0\cdot 5+3$.
Why might $r$ be the smallest element of $S$?
I’m not sure what you’re asking here. In the proof we simply let $r$ be the smallest element of $S$ and then prove that this number has the desired properties. Are you asking why we might think of trying this value of $r$? We already know that the remainder has to be in $S$, and we also know that it has to be less than $m$, so we want it to be small rather than large; it makes sense, therefore, to try the smallest member of $S$ to see whether it does the job.
The argument in step 8 is by contradiction: we suppose that $r\ge m$ and derive a contradiction, so that we can conclude that $r, just as we want. Recall that $r$ is the smallest member of $S$ and that $q$ has been chosen so that $n=qm+r$. Thus, $n-qm=r$, and since we’ve assumed that $r\ge m$, this implies that $n-qm\ge m$. Subtract $m$ from both sides of the inequality to get $n-qm-m\ge 0$, and then combine the two $m$ terms to get $n-(q+1)m\ge 0$. Now $q\in\Bbb N$, so $q+1\in\Bbb N$, and therefore $n-(q+1)m$ is one of the numbers in the list $(1)$. For convenience let $s=n-(q+1)m$; as we just saw, $s$ is in the list $(1)$. Moreover, we just showed that $s\ge 0$, so $s$ is actually in $S$. (Recall that $S$ is the set of non-negative members of the list $(1)$.)
But if you look back at the algebra that we just did, you can see that $s=n-(q+1)m=(n-qm)-m=r-m\;.$ Now recall that by hypothesis $m\in\Bbb N^+$, so $m>0$, and therefore $s=r-m. In other words, $s$ is a member of $S$ that’s smaller than $r$, contradicting our choice of $r$ as the smallest member of $S$. This contradiction shows that $r$ cannot be greater than or equal to $m$ and hence that $r.