If you have a group $G$ and $H \leq G$, the cosets of $H$ should partition $G$.
Suppose $G=\mathbb{Z}_2\times\mathbb{Z}_4$ and $H=\langle (0,1)\rangle = \{(0,0), (0,1),(0,2),(0,3)\}$. Then both $(1,1)+H$ and $(1,2)+H$ contain $(1,3)$. What am I doing that's dumb?
Nothing. Indeed
$$(1,2)+H=(1,1)+H\Longleftrightarrow (1,2)-(1,1)\in H$$
Remember: $\,xH=yH\Longleftrightarrow x^{-1}y\in H\,$ , or additively: $\,x+H=y+H\Longleftrightarrow -x+y\in H\,$