Why spherical coordinates is not a covering? Maybe this is an idiot question and I'm committing a trivial mistake. Let $\phi (\theta, \varphi) = (\cos \theta \sin \varphi, \sin \theta\sin \varphi, \cos \varphi)$ be the usual covering of the circle $\mathbb{S}^2 (1)$ by spherical coordinates
I always thought that polar coordinates could be seen as the picture, for instance, in here http://en.wikipedia.org/wiki/Spherical_coordinate_system (second picture). In this case it would be a covering space map. 

However when $\theta = 0$, the coordinates satisfies $(\sin \varphi, 0, \cos \varphi)$, which is a circle in the plane $y = 0$, but the angle $\theta$ is $0$ and $(-1, 0, 0)$ belongs to circle! However according to the image in the link $\theta$ should be $\pi$. So $\phi (0 \times (0, 2\pi)) \cap \phi (\pi \times (0, 2\pi)) \neq \emptyset$. Therefore I'm now not sure if $\phi$ is a covering space.
So, is $\phi$ a covering space map? And what's wrong with that $\theta$?
Rephrasing my question into one: in squares of what size $\phi$ is injective?
Thanks in advance.
EDIT
I've just realized that the map $\phi$ cannot be a covering. Since $\mathbb{S}^2(1)$ is already simply connected, $\mathbb{R}^2$ cannot be a covering space, otherwise it would be the universal covering.
Let $U_{a, b} = (a, 2\pi + a)\times(b, \pi + b)$. The function $\phi|_{U_{a, b}}$ is not an injection unless $a, b \in \mathbb{Z}$ (as noted by user86418). As discussed in the comments, I (and I think the other commenters) thought that $\mathbb{R}^2$ could be tessellated by the rectangles $U_{i, j}$ where $i, j \in \mathbb{Z}$. However this is not true, since this tessellation would not be induced by a group $G$ (otherwise $G \cong \pi_1 (\mathbb{S}^2) = 0$)
By making a more detailed analysis, it's possible to see that in the rectangle $R_{0,0} = \overline{U_{0, 0}}$, a vertical arrow pointing down is equivalent to an arrow pointing up (with the source) translated by $(\pi, 2\pi)$. More precisely, a point $(\theta, \varphi)$ is identified with a point $(\theta + \pi, 2\pi - \varphi)$. Together with this "action", there is the usual identification beetween  $(\theta, \varphi)$ and $(\theta + k2\pi, \varphi + l2\pi)$. 
In this case, the second action is given by $\mathbb{Z}^2$ (by translating by $2\pi$) and, in the quotient $X \cong \mathbb{R}^2/ \mathbb{Z}^2$, the first "action" turns in to a real action given by $\mathbb{Z}_2$. Therefore $\mathbb{S}^2 \cong X /\mathbb{Z}^2$. But, since $X \cong \mathbb{T}^2$ is a torus, this would imply that the sphere is a quotient of a torus by $\mathbb{Z}^2$. Is this correct?
So summarizing, the action on the torus by  the additive group $\mathbb{Z}^2$ is given by $1. (\theta, \varphi) = (\theta + \pi, 2\pi - \varphi)$.
So the questions are:Is $\mathbb{S}^2 \cong \mathbb{T}^2/ \mathbb{Z}^2$ as descripted above? Furthermore, for what open set $\phi$ fails to be a covering (i.e, when $\phi^{-1} (U)$ fails to be a disjoint union of isomorphic open sets)?
 A: No matter how you slice it (mathematicians' or physicists' conventions; latitude or colatitude, etc., etc.), spherical coordinates fail to be a covering map near the "poles", usually the points on the $z$-axis, where every longitude corresponds to a single point. (What time zone does the north or south pole of the earth lie in...?)
In your figure, fixing $r = 1$, you have
$$
(x, y, z) = \sigma(\theta, \varphi)
  = (\cos\theta \sin\varphi, \sin\theta \sin\varphi, \cos\varphi).
$$
(I've used "$\sigma$" for "spherical" rather than "$\phi$", in case my fingers mis-type "$\varphi$".)
The "non-covering" happens because if $k$ is an integer, then $\sigma(\theta, k\pi) = \bigl(0, 0, (-1)^{k}\bigr)$ for all real $\theta$. (As a consistency check, it's easy to verify the differential $D\sigma$ has rank one along the lines $\varphi = k\pi$.)
What's true is: For every integer $k$, the restriction $\sigma:\mathbf{R} \times \bigl(k\pi, (k+1)\pi\bigr) \to S^{2} \setminus \{(0, 0, \pm1)\}$ is a universal covering map for the sphere minus the north and south poles, a surface diffeomorphic to a cylinder. Each open rectangle of "width" $2\pi$ (and "height" $\pi$) in this strip is mapped diffeomorphically onto the sphere minus a closed half of a great circle joining the north and south poles.
(I haven't digested your edit carefully enough to know for certain what you mean by $T^{2}/\mathbf{Z}^{2}$, but in the usual reckoning, the integer lattice descends to a single point of the torus.)
Edit: In case a picture helps, the torus below is mapped, via its Gauss map, to the unit sphere. The "north" and "south" latitudes (in blue) are squeezed to points. Their complement wraps around the sphere twice; the shaded "inner" portion maps with degree $-1$, the unshaded "outer" portion maps with degree $+1$.

If $1 < R$, the torus can be parametrized by
\begin{align*}
\tau(\theta, \varphi)
  &= \bigl((R + \sin\varphi) \cos\theta, (R + \sin\varphi) \sin\theta, \cos\varphi) \\
  &= (R\cos\theta, R\sin\theta, 0)
  + (\cos\theta \sin\varphi, \sin\theta \sin\varphi, \cos\varphi) \\
  &= (R\cos\theta, R\sin\theta, 0) + \sigma(\theta, \varphi).
\end{align*}
The torus is a realization of $\mathbf{R}^{2}/(2\pi\mathbf{Z})^{2}$, and the parametrization $\tau$ is a universal covering map.
The summand $\sigma(\theta, \varphi)$, which coincides with spherical coordinates, is the value of the "outward-pointing" Gauss map at $\tau(\theta, \varphi)$. The Gauss map itself "discards" the term $(R\cos\theta, R\sin\theta, 0)$, and may be viewed as radially translating the longitudinal sections of the torus so they have a common axis (as if throttling a Slinky), so each blue point maps to a pole of the unit sphere.
Perhaps this makes clearer why the spherical coordinates map fails to be a covering map on any open set containing a point $\varphi = k\pi$ for some integer $k$.
A: There is nothing wrong with that notation (assuming the mathematical one, according to the Wikipedia link you included). Note that when $\theta=0$, increasing angle $\varphi$ goes from the Z axis to  the X axis. In other words, the negative X axis corresponds to $\varphi=\frac{3\pi}{2}$ and, hence, $(-1,0,0)$ does belong to the circle.
A: Honestly this is difficult to comment with no picture in the text, even if you linked one.
The problem is that you linked two of them...
And perhaps this is one of the answer you might have. 
In physics and mathematics the conventions are not the same. 
Don't try to set a value to $\theta$ and see wether it fits your understanding of the cartesian coordinates corresponding to it. 
Try to take any point on the sphere and see if you can have $(\theta,\varphi)$ that works for it. You will see that it does, whatever your convention, and that, in turn, $\phi$ is a covering space as you call it.
