This is claimed in various places. The problem seems to be with finding a free and transitive group action that has the fibers of $Mo$ as its orbits. I construct $Mo$ as $$ Mo = \mathbb{S}^1 \times \mathbb{R} / \sim$$ where $\sim$ is the equivalence relation $$ (t,x) \sim (t+2\pi,-x).$$
I am new to this topic and don't see why it would be impossible to construct an action. $(R,+)$ is certainly a Lie group.
On
Well, you said it yourself, if $Mo$ would have the structure of a principal $\mathbf R$-bundle over $S^1$ then any global section of it would prove that it is trivial. Take the zero section $0$. It is a global section and would trivialize the principal bundle. Contradiction since the total space $Mo$ is not homeomorphic to $S^1\times\mathbf R$.
However, if you take out the image of the zero section, you get a total space $Mo^\star$, say, that does admit the structure of a principal $\mathbf R^\star$-bundle! And as such, it is a nontrivial principal $\mathbf R^\star$-bundle.
Maybe what you are asking is: why is a principal bundle with a global section trivializable? Well, a principal bundle $G\to E\to B$ admits a free, transitive group action whose orbits are the fibers. If there is a principal section $s:B\to E$, then for every $b\in B$ we have $s(b)\in E$ and can map $g\to g\cdot s(b)$. This map provides a trivialization of $E$ by identifying $$e = g\cdot s(b) \leftrightarrow (g,b) $$ So you see, because the Mobius band has a global section --- the zero section --- if it admitted a free and transitive $\mathbb{R}$ action, that global section could be "smeared" into a trivialization. In this case, the $\mathbb{R}$ action must be addition, so to visualize this fact, ask yourself: if a rubber band is stretched around the core of a Mobius strip, can I roll it one centimeter off the core?