For your first question: any non-linear homography will do. Just take the foliation given by $dy=0$, for which the "constant" functions $x\mapsto (x,c)$ are leaves (notice that it is not actually constant in the sense you probably meant in the question), and then apply any homographic change of coordinates in the $y$-variable.

For your second question: no it is not true. By the uniformization theorem, the universal covering of a leaf $L$ is either the Riemann sphere $\bar {\mathbb{C}}$, the complex line $\mathbb C$ or the unit disc $\mathbb D$. The latter is the generic situation for generic polynomial foliations. Now, any entire, surjective map $f : \mathbb C\to L$ would realize a holomorphic cover. Because $\mathbb C$ is simply connected, it means that $L$ has $\mathbb C$ for universal cover.

For your first question: $dy=y^2$ can be integrated by quadratures. The solutions are homographies.

For your second question: no it is not true. By the uniformization theorem, the universal covering of a leaf $L$ is either the Riemann sphere $\bar {\mathbb{C}}$, the complex line $\mathbb C$ or the unit disc $\mathbb D$. The latter is the generic situation for generic polynomial foliations. Now, any entire, surjective map $f : \mathbb C\to L$ would realize a holomorphic cover. Because $\mathbb C$ is simply connected, it means that $L$ has $\mathbb C$ for universal cover.