I have a question about a very basic thing. Let $M$ be a smooth manifold and suppose that we have a smooth map $g:M\to {\rm SO}(D)$. Let $\rho : {\rm Spin}(D)\to {\rm SO}(D)$ be the covering map. Since $\ker \rho = \{\pm 1\}$ at each $x\in M$ the equation $$\rho(\tilde{g}(x))=g(x)\tag{1}$$
has two solutions. So choosing one solution at each $x\in M$ this defines a function $\tilde{g}:M\to {\rm Spin}(D)$. Intuitively it is very clear to me that we can choose the solutions so that $\tilde{g}$ is smooth as well, but I fail to see how to prove it. I mean this is something that is probably extremely basic, but I'm really missing it. So how could we argue that it is always possible to pick the solutions of (1) at each $x\in M$ such that the resulting function $\tilde{g}$ is smooth?
It's possible to choose a continuous lift $\tilde g$ locally, but it may not be possible to do so globally.
As a simpler counterexample, consider the circle $S^1$ as the set of unit complex numbers $e^{i\theta}\in\mathbb{C}$. The map $c(z)=z^2$ is a double covering map $S^1\to S^1$. The identity $\operatorname{id}:S^1\to S^1$ is of course continuous, but there is no continuous map $f:S^1\to S^1$ such that $c\circ f=\operatorname{id}$. This can be proven using fundamental groups.
The same is true in your case. If we take $M=S^1$ and choose a map $g:S^1\to SO(D)$ which is not homotopic to a constant map, then there will be no continuous lift $\tilde{g}$ satisfying $\rho\circ\tilde{g}=g$.