When checking the transition maps for differentiability in order to determine if a manifold is differentiable, do we fill in any removable singularities (i.e. simplify the function composition before differentiating)? Or do these removable singularities make a difference?
The example I'm thinking about is the manifold $\mathbb{R}$ with the single chart $x\mapsto\sqrt[3]{x}$.
If you don't ignore removable singularities then this atlas is only $C^0$ since it's neither compatible with itself nor (by the chain rule) any other atlas beyond $C^0$.
But if you do ignore removable singularities this leads to odd behavior, such as a smooth manifold with a chart that isn't even differentiable.
I think you're worrying about something that's not happening. What you've done is given $\Bbb R$ a differentiable structure with precisely one chart, the map $\varphi: x \mapsto x^{1/3}$. Now the compatibility demand is that given two charts $\varphi, \psi$ (defined on all of $\Bbb R$ for convenience of notation), the map $\varphi \circ \psi^{-1}$ is smooth. In particular, note that we don't take derivatives until after we've passed to transition functions.
But given any chart $\varphi$ (in particular yours!), $\varphi \circ \varphi^{-1} = \text{Id}$, no matter what, no singularities, etc. Every chart is compatible with itself.
(I suspect the upset stomach comes because you know that $x^{1/3}$ is not differentiable at the origin. This would cause trouble if, say, we were trying to show that this chart was compatible with the standard (identity) chart - it's not! But we don't need to take the derivative of $x^{1/3}$ itself at any point in the process if we're just trying to show this chart is compatible with itself.)