The quotient space $Y = X / \sim$ as a set is just the set of equivalence classes of $X$ under $\sim$, so the set $\{ [x]: x \in \mathbb{R} \} $ in your case.
The equivalence class of a number $x$ is just (in your case) the set $\{ x+n : n \in \mathbb{Z} \}$. Now we need a topology. The standard topology that we take on $Y$ is all subsets $O$ of $Y$ (where points in $Y$ are "really" subsets of $X$, the equivalence classes) such that $q^{-1}[O]$ is open in the topology of $X = \mathbb{R}$. Here $q$ is the map that sends $x$ to its class $[x]$ in $Y$, the so-called quotient map. This is called the quotient topology on $Y$, and as you see it assumes you have a topology on $X$ already, and we give $Y$ the largest topology possible to still have $q$ continuous. (The smallest one would always be the indiscrete topology, which is not very interesting, hence the other "natural" choice.)
Now, if we have a function $f$ from $Y$, the quotient space in the quotient topology, to any space $Z$, then $f$ is continuous iff $f \circ q$ is continuous as a function from $X$ to $Z$: one way is clear, as the composition of continuous maps is continuous, and for the other side, if $O$ is open in $Z$, by definition $f^{-1}[O]$ is open in $Y$ iff $q^{-1}[f^{-1}[O]]$ is open in $X$, and this set equals $(f \circ q)^{-1}[O]$ which is open, as by assumption $f \circ q$ is continuous.
Now, consider the map $f$ that sends the class $[x]$ to the point $e^{2\pi ix}$ in $\mathbb{S}^1$, the unit circle.
This is well-defined: if $x'$ were another representative of $[x]$, then $x \sim x'$ and thus $x - x'$ is an integer and so $f(x') = f(x)$.
It is continuous, as $f \circ q$ is just the regular map sending $x$ to $e^{2\pi ix}$, and this is even differentiable etc.
It is clearly surjective and injective because the only way $[x]$ and $[y]$ will have the same value is when $2 \pi ix - 2 \pi i y$ is an integer multiple of $2 \pi i$, which happens iff $x - y$ is an integer.
One can also check that $q[X] = q[[0,1]]$ and by continuity of $q$ we have that $Y$ is compact. This makes (with $\mathbb{S}^1$ Hausdorff) the map $f$ a homeomorphism, by standard theorems.
We could also have achieved this as the quotient of $[0,1]$ under the equivalence relation that has exactly one non-trivial class, namely $\{0,1\}$. This is more intuitive, as we then glue together (consider as one point) just the points $0$ and $1$, and this geometrically gives a circle. In your example (which is a nice so-called covering map, and a group homomorphism as well) we glue a lot more points together, but all classes are now similar: just shifted versions of a point by an integer. We sort of wrap the interval $[0,1)$ infinitely many times over itself.
Let $X$ be a topological space, and let $A\subseteq X$. $A$ defines the following equivalence relation $\overset{A}\sim$ on $X$: for $x,y\in X$, $x\overset{A}\sim y$ iff either $x=y$, or $\{x,y\}\subseteq A$. The quotient space $X/A$ is defined to be the same as the quotient space $X/\overset{A}\sim$.
Added: To put it a little differently, the partition of $X$ induced by $\overset{A}\sim$ is $$\{A\}\cup\big\{\{x\}:x\in X\setminus A\big\}\;.$$ The topology of of the quotient space is defined as follows: $V\in X/A$ is open iff $\pi^{-1}[V]$ is open in $X$, where $\pi:X\to X/A$ is the obvious quotient map.
As Alex Becker mentions in the comments, this notion of quotient is somewhat different from the notion used when some algebraic structure is present, as in the case of quotients of groups, rings, vector spaces, etc.
Best Answer
Quotient spaces are not always very easy to visualize. What's worse, there is no uniform method that applies to let you visualize them all. Like most things (in and out of mathematics), you simply have to experience a lot of quotient spaces to learn how to understand, and sometimes visualize, them.
For this quotient space $X^*$, first let's look at some important subspaces. First let's ignore the two points $(0,-1)$ and $(0,+1)$. Notice that $(\mathbb R - \{0\}) \times \{-1\}$ and $(\mathbb R - \{0\}) \times \{1\}$, which one can think of as "two copies of the line minus the origin", are being identified into one single "line minus the origin", a subspace of $X^*$ that I'll denote $L$.
Next, $X^* - L$ has just two other points, which I'll denote $0^- = \{(0,-1)\}$ and $0^+ = \{(0,+1)\}$. The union $L \cup \{0^-\}$ is homeomorphic to $\mathbb R$, and the union $L \cup \{0^+\}$ is also homeomorphic to $\mathbb R$.
So, one can think of $X^*$ as a line with two origins, which is what people actually call this beast. If you want to try to visualize it, imagine that you are simply looking at the actual real line, except that when you try to focus on the origin you have double vision. That's about the best one can say for this example.