It seems clear that the unit square curve $(0,0)-(0,1)-(1,1)-(1,0)-(0,0)$ can be parametrized by two charts $\varphi_1$ and $\varphi_2$ over $R$, that overlap in a differentiable way (e.g. each chart maps one side and 2/3 of the adjacent sides in the square to $R$). So, this curve endowed with these charts would be a differential manifold? isn't it somewhat counter intuitive since the square has corners?
Even more strikingly, one could consider a strictly increasing function from $R$ to $R$ nowhere differentiable. Obviously, this define a single chart from its graph to $R$. Hence this graph with this chart would be a differential manifold?
A differentiable manifold is a topological manifold (= second countable Hausdorff locally Euclidean space) $M$ with a differentiable structure. This is a structure component which comes in addition to the topology of $M$. Given a topological manifold $M$, we can usually endow it with various distinct differentiable structures (although there exist examples where no differentiable structure on $M$ exists).
As you say, the square can be endowed with a differentiable structure. However, it is not a differentiable submanifold of $\mathbb R^2$ (the corners prevent this).
Given any homeomorphism $h : \mathbb R \to \mathbb R$, you can endow $\mathbb R$ with the atlas $\mathcal A(h) = \{ h \}$. This is a differentiable atlas since it has no nontrivial transition functions. If one of $h, h^{-1}$ is not differentiable, then the differentiable structure $\mathcal D(h)$ generated by $\mathcal A(h)$ differs from the standard differentiable structure $\mathcal D(id)$. However, the map $h : (\mathbb R, \mathcal D(h)) \to \mathbb R = (\mathbb R, \mathcal D(id))$ is a diffeomorphism so that you do not get something exotic by using $\mathcal D(h)$.