Why Does a Torus Require Four Patches to Cover?

The book I'm reading implies at least four coordinate patches (simple surfaces) are needed to cover a torus. Consider the following mapping:

$$\textbf{x}(u^1, u^2) = \begin{bmatrix} ((2+cos(u^1))cos(u^2) \\ ((2+cos(u^1))sin(u^2) \\ sin(u^1) \end{bmatrix} $$

where $u^1, u^2 \in(-\pi, \pi)$.

The issue here is we are missing the points when $u^1,u^2 =\pm\pi$. Geometrically, we are missing the points on a circle lining the inside of the torus, and another circle orthogonal to the aforementioned one, that does not line the inside of torus but rather goes about the "tube" part. Why can't we then just use the same mapping as above but with $u^1, u^2\in(0, 2\pi)$?

There is an atlas for the $2$-torus $T$ with three charts. If we view $T$ as the result of doing identifications as usual on a square $Q$, then the three open sets are

• the complement of the image in $T$ of the horizontal sides of $Q$,

• the complement of the imagfe in $T$ of the vertical sides of $Q$, and

• a small disk in $T$ centered at the image of the vertices of $Q$.

These are two annuli and a disk.

I am pretty sure you cannot do this with with two charts. If one adds the requirement that the domains of the charts are contractible, this follows from the theory of Lusternik–Schnirelmann category and its lower bound in terms of cup length. But of course, two of my open sets above are not contractible...