If for each $X$ a smooth manifold we define a function $f: X → \mathbb{R}$ to have a derivative of order $n>0$ if for each parameterization $\phi: U \to V\subseteq X$, if the composition $f\circ \phi$ are differentiable from any order $≤ n$, how can I show that $f$ is continuous ?
This is not directly due to the fact that $f$ has a derivative of order $n$?
In the context of smooth manifolds, parametrizations (or rather, charts) are usually maps from open subsets of the manifold to some euclidean space, but you can just take the inverse and formulate it the way you did.
It is a common exercise to show that the charts are diffeomorphisms, so in particular $\phi$ is continuous, biyective, and its inverse is continuous.
Since $f\circ \phi:\Bbb{R}^n\to\Bbb{R}$ is differentiable, it is continuous by the theorems regarding differentiability on euclidean spaces. Since $\phi^{-1}$ is continuous, so is $(f\circ\phi)\circ\phi^{-1}=f\circ (\phi\circ \phi^{-1})=f$.