Is the parametrization of the flat torus $$\mathbb{T}^2 = \mathbb{E}^2/\mathbb{Z}^2$$ just the set of charts $$(\pi,U_{ij})\quad \pi(x)=x\quad U_{ij}=(i,i+1)\times(j,j+1)\quad i,j\in\mathbb{Z}^2$$
EDIT: I mistook the quotient group notation for the set minus notation. My revised answer is then
$$(\pi,U)\quad \pi(x)=x\quad U=[0,1)\times [0,1)$$
I omitted top and right edges, I'm not sure if this is correct?!
EDIT:
$$(\pi,U)\quad \pi_1(x)=x\quad U_1=(0,1)\times (0,1)$$ $$(\pi,U)\quad \pi_2(x)=x+1/4\quad U_2=(-1/2,1/2)\times (-1/2,1/2)$$
No, those charts are not enough.
First, each of the charts $(\pi,U_{ij})$ covers exactly the same orbits, and so they cover exactly the same points in $\mathbb{T}^2$. The reason this is true is because for each $i,j,i',j' \in \mathbb{Z}$ the translation $(x,y) \mapsto (x+i'-i,y+j'-j)$ takes $U_{ij}$ to $U_{i'j'}$. So this list of charts is redundant. You can throw away all but one of them, keeping $U_{0,0}$ say. The chart $U_{0,0}$ covers exactly the same point of $\mathbb{T}^2$ as are covered by your whole collection of charts.
However, that chart does not cover the orbit of the point $(0,0)$ itself, nor does it cover the orbit of any of the points of the form $(m,y)$ where $m$ is an integer, nor of the form $(x,n)$ where $n$ is an integer. The only points whose orbits are covered by your charts are those points of the form $(x,y)$ where neither $x$ nor $y$ is an integer.
Added to address edits: The trouble with the new charts that you propose in your new edit is that a chart in an atlas for $\mathbb{T}^2$ must be a homeomorphism from an open subset of $\mathbb{R}^2$ to an open subset of $\mathbb{T}^2$. But $[0,1) \times [0,1)$ is not an open subset of $\mathbb{R}^2$.