Let $M$ be a manifold with a global chart, and let $N$ be a non-empty compact manifold.
Let $f\colon M\to N$ be a function such that $f(M)=N$.
To study the smoothness of $f$, I have to choose a point $x\in M$, a chart $U$ of $M$ around $x$ and a chart $V$ around $f(x)$ such that $f(U)\subseteq V$. Next, I have to study the local representative of $f$ with respect to the selected charts.
In my case, the only possibility of a chart around any point of $M$ is the global one. Moreover, there exists no chart $(V,\psi)$ of $N$ such that $f(M)\subseteq V$ since $N$ can't be covered by a single chart.
Why is this argument wrong?
Maybe the point is that equivalent classes of atlases give the differentiable structure of $M$. In particular, if $((V_i,\psi_i))_{i=1}^n$ is an atlas of $N$ and $(M,\varphi)$ is the global atlas of $M$, then $((f^{-1}(V_i),\varphi\restriction f^{-1}(V_i))_{i=1}^n$ is an equivalent atlas of $M$ such that $f(f^{-1}(V_i))=V_i$.
You've pretty much answered your own question in the last paragraph: simply restrict $f$ to the chart $(f^{-1}(V_i), \phi \mid f^{-1}(V_i))$.
This is part of the reason underlying the concept of a "maximal" atlas. The global chart is not a maximal atlas all by itself. But when formulating the maximal atlas that contains the global chart, you are forced to include all other charts formed from open subsets of that global chart, in particular the chart $(f^{-1}(V_i), \phi \mid f^{-1}(V_i))$.