similar setting but with a fibre preserving simply transitive right action
Well, for any group $G$ acting on a set $P$, if you look consider the fibers of the quotient set $P/G$, these are by definition the orbits of the group. So, transitivity of the action on the fibers is automatic, meaning one only needs to impose freeness, if you decide to introduce the concept this way.
Other times, people might introduce the definition by allowing for a possibly different base manifold $B$ which is not $P/G$ set theoretically. But really, the choice of set is not what matters but rather how the group action $G$ plays with the projection map $P\to B$, i.e what we require is, roughly speaking, a surjective map $\pi:P\to B$ such that the action of $G$ on $P$ restricts, for each $b\in B$, to a transitive action on the fiber $\pi^{-1}(\{b\})$ (which is equivalent to fixing a point $p$ in this fiber and saying this fiber equals the orbit of $p$ under $G$).
So, as far as definitions are concerned, they may look different, but that is only superficial.
Question 1.
You do not need compactness of the group. Take a look at the following smooth orbit-stabilizer theorem. The definition of a principal bundle in the first sense requires that
- $G$ be a Lie group which acts smoothly on the right on a smooth manifold $P$.
- The orbit manifold $P/G$ must exist, i.e there must exist a smooth structure on $P/G$ such that the underlying topology is the quotient topology for the projection $\pi:P\to P/G$, and this map $\pi$ must be a smooth submersion.
- $G$ must act freely on $P$.
What the lecturer doesn’t emphasize is the second bullet point. Just FYI: A necessary and sufficient condition (I think this is called Godement’s criterion) is that the orbit relation
\begin{align}
\mathcal{R}:=\{(x,y)\in P\times P\,:\, \text{$x,y$ lie in the same $G$-orbit}\}
\end{align}
must be a closed subset and a smooth embedded submanifold of $P\times P$. This is somewhat difficult to check in practice, so we typically settle for a simpler sufficient condition, namely that $G$ acts freely and properly (and properness will trivially be satisfied if $G$ is compact).
So, now by closedness of the orbit relation, it implies each orbit is closed (it is the preimage of a singleton (which is closed since we have a Hausdorff topology on $P/G$) under the projection $\pi:P\to P/G$), so by the result in the link, we have that for each $p\in P$, the orbit $\mathcal{O}_p$ is an embedded submanifold of $P$ diffeomorphic to the quotient $G/\text{Stab}(p)$, and this latter set is $G/\{e\}$ since the action is free, so trivially diffeomorphic to $G$.
Question 2.
First what you do is show that locally there always exist smooth sections of $\pi:P\to P/G$. This is a simple consequence of the fact that $\pi$ is a surjective submersion. So, fix a smooth local section $\phi:U\subset P/G\to \pi^{-1}(U)\subset P$. Since we have a free action, we see that the map $\Phi:U\times G\to \pi^{-1}(U)$, $(x,g)\mapsto \phi(x)\cdot g$ is injective (surjectivity is obvious), so it is bijective, and it is certainly smooth and it preserves base points. All that remains is showing that the inverse is smooth. This follows by an inverse-function theorem type of argument so I omit it.
So, $\Phi$ is now a diffeomorphism $U\times G\to \pi^{-1}(U)$ which preserves the base points. On the domain, we have an obvious right action of $G$ namely $(x,s)\cdot g:=(x,sg)$ where $sg$ is the usual group product. On the target, $\pi^{-1}(U)$ also has a natural action of $G$ by restricting the one on $P$. Observe that we have for each $(x,s)\in U\times G$ and $g\in G$,
\begin{align}
\Phi((x,s)\cdot g):=\Phi(x,sg):=\phi(x)\cdot (sg) = [\phi(x)\cdot s]\cdot g=\Phi(x,s)\cdot g,
\end{align}
so we have an equivariance condition satisfied by $\Phi$.
To summarize starting from the first definition, the way you get an equivariant bundle trivialization is by taking a local section (which is a way to ‘lift’ the base manifold into the bundle), and then you act by $G$ to move up and down the fiber (freeness and fiberwise-transitivity then make this all bijective).
