Is it correct to say that a loop is null homotopic iff it can be contained in a single (simply connected) chart? That is, for a loop $\gamma(t)$ based at $p$, there's a chart $(\phi, U)$ such that $U$ is simply connected and $\gamma$ is within the domain of $\phi$.
It seems like a reasonable enough assumption, since the contraction of $\gamma$ to a point, and possibly slight deformations of the surface generated to be the correct number of dimensions, would be homeomorphic to a ball, but I am not quite sure how to prove this.
The statement a loop is null-homotopic iff it is contained in a simply-connected chart is wrong. For example consider the following loop $\gamma(t) = e^{it}$ if $0 \leq t \leq 2 \pi$ and $\gamma(t) = e^{-it}$ if $2\pi \leq t \leq 4 \pi$. This loop is covering all $S^1$ but is not contained in a single chart since $S^1$ can't be homeomorphic to $\mathbb R^1$. It is clearly null-homotopic since by definition this is $a - a = 0$ where $a$ is the positive generator of $\pi_1(S^1)$.
The statement a loop is null-homotopic iff it is homotopic to a loop contained in a simply-connected chart is true but a bit tautological : by hypothesis the loop is homotopic to the constant loop which is certainly contained in a simply-connected chart.