The differences between $\mathbb{R}/ \mathbb{Z}$ and $\mathbb{R}$ 
The cosets of $\mathbb{Z}$ in $\mathbb{R}$ are all sets of the form $a+\mathbb{Z}$, with $0 ≤ a < 1$ a real number. Adding such cosets is done by adding the corresponding real numbers, and subtracting 1 if the result is greater than or equal to 1. -- Examples of Quotient Group, Wiki

I cannot figure out the differences between $\mathbb{R}/ \mathbb{Z}$ and $\mathbb{R}$. Besides, "subtracting 1 if the result is greater than or equal to 1", what does "the result" mean here? Why do we need to subtract 1? I was wondering what is the background of $\mathbb{R}/ \mathbb{Z}$. 
 A: Every real number has two parts: an integer part and a decimal part. In other words, we can write every $x \in \mathbb{R}$ as a sum of elements $n_{x} \in \mathbb{Z}$ and $u_{x} \in [0,1]$. Let us write $x = n_{x} + u_{x}$. Now, we can use this to define an equivalence relation on $\mathbb{R}$: we say that $x \equiv y$ if $u_{x} = u_{y}$, so that their decimal parts are the same. For a given element $x \in \mathbb{R}$, the equivalence class $[x]$ is just $u_{x} + \mathbb{Z}$.
The quotient group $\mathbb{R}/\mathbb{Z}$ arises by "collapsing" together these equivalence classes. We are thus left with something akin to the unit interval $[0,1]$, as the rest of $\mathbb{R}$ is made of shifts of the unit interval by $\mathbb{Z}$. However, since $0$ and $1$ have the same decimal expansion, they are equal to each other. What we get, topologically, is a line with the ends tied together: a circle. Thus $\mathbb{R}/\mathbb{Z} \cong S^{1}$, and hence it is compact. 
Moreover, we need to be careful in adding equivalence classes. Clearly, $[0.25] + [0.25] = [0.5]$. However, $[0.75] + [0.75] = [1.5]$. Since we working with the unit interval, we subtract one, to get $[0.5]$. The equivalence class is the same, only now we are representing it by an element in $[0,1)$. 
A: $(a+\mathbb Z)+(b+\mathbb Z)$ is found by adding $a$ and $b$, the result of which is $a+b$.  If $a+b<1$, then $(a+\mathbb Z)+(b+\mathbb Z)=(a+b)+\mathbb Z$.  If $a+b\geq 1$, then $(a+\mathbb Z)+(b+\mathbb Z)=(a+b-1)+\mathbb Z$.
But this is only if you follow the stated convention of only listing representatives from $[0,1)$.  The fact is, $(a+b)+\mathbb Z$ and $(a+b-1)+\mathbb Z$ are different names for the exact same set, so you don't really need to subtract $1$.
A: The elements of $\mathbb{R}$ are real numbers. The elements of $\mathbb{R}/\mathbb{Z}$ are sets of real numbers, pairwise disjoint, where each set contains reals that differ from each other by integer distances, and the set contains all reals that differ from the reals in the set by integer distances.
So one of the elements of $\mathbb{R}/\mathbb{Z}$ is
$$\{ \ldots, -2, -1, 0, 1, 2, 3, \ldots\}$$
Another element of $\mathbb{R}/\mathbb{Z}$ is
$$\{\ldots, \pi-3, \pi-2, \pi-1, \pi, \pi+1, \pi+2,\ldots\}$$
and so on. Every real number is in one and only of the elements of $\mathbb{R}/\mathbb{Z}$ (which, remember, are sets of real numbers).  
(The elements of $\mathbb{R}/\mathbb{Z}$ are the equivalence classes of real numbers under the equivalence relation "$x\sim y$ if and only if $x-y\in\mathbb{Z}$").
Now, the set $\mathbb{R}/\mathbb{Z}$ can be made into a group as well; that is, we can define an "addition of classes". One way to define this "addition of classes" is to first give a "name" to each element of $\mathbb{R}/\mathbb{Z}$. Since every one of the sets contains one, and only one, real number in $[0,1)$, we will represent the set that contains the real number $r\in[0,1)$ by writing $[r]$. So the first set I wrote above is called $[0]$, the second set I wrote above is called $[\pi-3]$, and so on.  
(Added. Given a real number, how can you tell what element of $\mathbb{R}/\mathbb{Z}$ it is in? Remember that $\lfloor x\rfloor$ is defined to be the largest integer $n$ such that $n\leq x$. The "fractional part of $x$" is defined to be $ x-\lfloor x\rfloor$. It is not hard to check that $x\in[x-\lfloor x\rfloor]$.)
Now that every set has a name, we define addition of classes. The way we are going to define addition (which I will write $\oplus$ to distinguish it form the addition of real numbers) is as follows: if $[r]$ and $[s]$ are two classes, then $[r]\oplus [s]$ is:
$$[r]\oplus [s] = \left\{\begin{array}{ll}
\ [r+s] &\text{if }0\leq r+s\lt 1\\
\ [r+s-1] &\text{if }1\leq r+s.
\end{array}\right.$$
Because $0\leq r,s\lt 1$, then $0\leq r+s\lt 2$, so one and only one of those situations will happen, and the result will always be a "proper name" for an element of $\mathbb{R}/\mathbb{Z}$ (that is, the answer is of the form $[a]$ with $0\leq a\lt 1$). 
Then one can show that this makes $\mathbb{R}/\mathbb{Z}$ into a group.
What the text you are quoting is doing is that it is writing "$r+\mathbb{Z}$" where I wrote "$[r]$" above. The reason it does this is that this is the standard notation when performing the kind of construction that is being discussed. This is covered in any book on abstract algebra that discusses groups.
A: I wrote some words about the algebraic structure on $\mathbf R/\mathbf Z$, but I think that's been treated very well in the other answers. By the tags I feel that we should also talk about the topological and analytic aspects of this group, so here's a start in that direction. I certainly wish that I knew more about this.
$\mathbf R/\mathbf Z$ is often called the circle group, because if we view $S^1$ as lying inside of the complex plane then the map $F\colon\mathbf R \to S^1$, $F(x) = e^{2\pi ix}$ induces a bijection $\mathbf R/\mathbf Z \to S^1$ which is both an isomorphism of groups and a homeomorphism of topological spaces (if we place the quotient topology on $\mathbf R/\mathbf Z$); in particular, $\mathbf R/\mathbf Z$ is naturally a compact space.
$\mathbf R/\mathbf Z$ is a nice home for periodic functions, since if $f\colon \mathbf R \to \mathbf R$ is such that $f(x + n) = f(x)$ for all $x \in \mathbf R$ and $n \in \mathbf Z$, then $f$ descends to a map $\mathbf R/\mathbf Z \to \mathbf R$. Moreover, Pontryagin duality says that Fourier analysis on the circle is related to functions on a certain "dual group", and the dual group of the circle is $\mathbf Z$, which is discrete! On the other hand, $\mathbf R$ is its own dual.
As for differential operators, that's a large subject and I can only sketch the situation and give Wiki references. One can view $\partial/\partial x$ as a vector field on $\mathbf R$. Each $a \in \mathbf R$ induces a translation diffeomorphism $L_a\colon \mathbf R \to \mathbf R$, $L_a(x) = x + a$, and the field $\partial/\partial x$ is called left-invariant if it is unchanged under the corresponding pushforwards: $L_{a*}(\partial/\partial x) = \partial/\partial x$.
In general, given a smooth map of manifolds $F\colon N \to M$ a vector field on $N$ may not correspond to a vector field on $M$. But you can check that in this case we can define a vector field on $S^1$, which people usually call $\partial/\partial\theta$, by $(\partial/\partial\theta)_{F(P)} = F_*(\partial/\partial x)_P$.
A: One difference between ${\bf R}/{\bf Z}$ and $\bf R$ is that $${1\over2}+{1\over2}=0$$ in the former but not in the latter. 
EDIT: Perhaps it would be more precise to write $$\left({1\over2}+{\bf Z}\right)+\left({1\over2}+{\bf Z}\right)=0+{\bf Z}$$ but $${1\over2}+{1\over2}\ne0$$ 
Perhaps even better simply to say that if $x+x$ is the identity element in $\bf Z$ then $x$ is itself the identity element of $\bf Z$, but in ${\bf R}/{\bf Z}$ there is an $x$ other than the identity such that $x+x$ is the identity. 
