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.
Define a map $f$ from $I^2$ to $X$ which sends $(x,y)$ to $(x+1)\cdot (\textrm{cos}(2\pi yi),\ \textrm{sin}(2\pi yi))$. To make $I^2$ into a torus you have to glue opposite sides together, i.e. $(x,0)\sim_T(x,1)$ and $(0,y)\sim_T(1,y)$ for $x,y\in I$. The torus $T^2$ is then the quotient space $I^2/\sim_T$. Show that points are identified in $I^2$ if and only if their images are identified in $X$. Thus $f$ induces an injection $\hat f$ between $T^2$ and $X/\sim$. Since $f$ is surjective, $\hat f$ is bijective. Now you have to prove that $f$ is an identification mapping, that is a set $U\subseteq X$ is open iff its preimage $f^{-1}(U)$ is open. Finally, you should remember how this implies the openness of $\hat f$.
Best Answer
I'll consider $S^1$ as the unit circle in $\Bbb{R}^2$, and hence $S^1\times S^1$ as a subset of $\Bbb{R}^4$.
For every point $(x,y)$ on the annulus $A$ there are unique $r\in[1,2]$ and $\theta\in[0,2\pi)$ such that $(x,y)=(r\cos\theta,r\sin\theta)$, and the map $$A/\sim\ \longrightarrow\ S^1\times S^1:\ (x,y)\ \longmapsto\ \left(\left(\cos\theta,\sin\theta\right), \left(\cos(2(r-1)\pi),\sin(2(r-1)\pi)\right)\right),$$ is a homeomorphism.