For the first question - yes, at least if you suppose $P$ is a smooth manifold and, say, $G$ is a Lie group. For the principal $G$-bundles with your definition, just as with the regular one, being trivial is the same as admitting a section (in the section is $s$, map $s(x)$ to ($x$, unit of $G$), and use the action to define the rest of the trivializing map). Now to construct a section locally, near $x_0$ take any $s(x_0)$ in the fiber above $x_0$. Pick an auxiliary Riemann metric near take the orthogonal subspace to the tangent of the fiber at $s(x)$. Exponential map will give you a section locally.
In other categories you would need to construct the section $s$ in a different manner. I think this can be done in the category of topological manifolds. Not sure about more general cases, but it would seem ok. Maybe for CW complexes you can go cell by cell?
No, if a vector bundle is trivial as a smooth fiber bundle, then it is also trivial as a vector bundle. In fact, a more general result is true: if any two smooth vector bundles are isomorphic as smooth fiber bundles, then they are isomorphic as vector bundles.
[This proof is a slightly modified version of the one I originally posted, adapted to prove the more general result. For reference, my original proof is reproduced below.]
The key idea is that every smooth fiber bundle with a global section has a vector bundle associated with it, namely the pullback of the vertical tangent bundle along the section; and if two fiber bundles are isomorphic, then so are their pullback vertical bundles. On the other hand, if a fiber bundle also happens to have the structure of a smooth vector bundle, then the pullback vertical bundle is naturally isomorphic to the vector bundle itself.
In more detail, here's how it works. Suppose first that $\pi\colon E\to M$ is a smooth fiber bundle with $k$-dimensional fibers. There is a rank-$k$ vector bundle $T^V E\to E$, called the vertical tangent bundle, whose fiber at a point $p\in E$ is the tangent space to the fiber $E_{\pi(p)} = \pi^{-1}(\pi(p))$: in other words, $T^V_pE = T_p(E_{\pi(p)}) = \ker d\pi_p$.
If $E$ has a global section $\sigma\colon M\to E$, we let $E_\sigma\subset E$ be the image of the global section, which is a smooth embedded submanifold diffeomorphic to $M$. The
restriction $T^V\!E|_{E_\sigma}$ is a rank-$k$ vector bundle over
$E_\sigma$, which we denote by $E^V\to E_\sigma$. It can be considered as the
subset of $TE$ consisting of all vertical vectors over points of $E_\sigma$.
Now suppose $\pi'\colon E'\to M$ is another smooth fiber bundle that is isomorphic over $M$ to $E$ (as a smooth fiber bundle). Thus there is a smooth
diffeomorphism $\Phi\colon E\to E'$ covering the identity map of $M$. We obtain a global section $\sigma'=\Phi\circ\sigma\colon M\to E'$, and we can perform the same construction on $E'$ to yield a vector bundle $E^{\prime V}\to E'_{\sigma'}$. Because $\Phi$ is a bundle map, the global differential $d\Phi\colon TE\to TE'$ restricts to a bundle isomorphism from $E^V$ to $E^{\prime V}$ covering the diffeomorphism $\Phi|_{E_{\sigma}}\colon E_{\sigma} \to E_{\sigma'}'$.
On the other hand, if $E\to M$ is a smooth vector bundle and $\sigma\colon M\to E$ is any global section (for example, the zero section), we can construct the vector bundle $E^V\to E_{\sigma}$ as before. But in this case,
for each point $q\in M$, the fiber $E_q\subseteq E$ is a vector space, and
the fiber $E^V_{\sigma(q)}\subseteq E^V$ is the
tangent space to $E_q$ at $\sigma(q)$. Each tangent space to the
finite-dimensional vector space $E_q$ is canonically isomorphic to the vector space $E_q$ itself; the isomorphism is given by sending an element $v\in E_q$ to the derivation $D_v\colon C^\infty(E_q) \to \mathbb R$ defined by $D_v(f) = (d/dt)|_{t=0} f(\sigma(q)+tv)$.
Let $\alpha\colon E \to E^V$ be the map whose
restriction to each fiber $E_q\subseteq E$ is the canonical isomorphism
$E_q\to T_{\sigma(q)}(E_q) = E^V_{\sigma(q)}$. Then $\alpha$ is a vector bundle isomorphism covering the diffeomorphism
$\sigma\colon M\to E_{\sigma}$, provided it is smooth.
In a neighborhood $U$ of any point of $M$, there is a local vector bundle trivialization $\Psi\colon \pi^{-1}(U)\to U\times \mathbb R^k$. Its differential restricts to a smoooth local trivialization $d\Psi|_{(\pi^V)^{-1}(U)}\colon (\pi^V)^{-1}(U) \to U\times \mathbb R^k$. Unwinding the definitions shows that the map $d\Psi\circ\alpha\circ \Psi^{-1}\colon U\times \mathbb R^k\to U\times \mathbb R^k$
has the form $d\Psi\circ\alpha\circ \Psi^{-1}(q,v) =(q,v)$.
Since $\Psi$ and $d\Psi|_{(\pi^V)^{-1}(U)}$ are diffeomorphisms, this shows that $\alpha$ is smooth in a neighborhood of each point.
Putting this all together, if $E\to M$ and $E'\to M$ are smooth vector bundles that are isomorphic over $M$ as smooth fiber bundles, then we have a composition of vector bundle isomorphisms
$$
E\overset{\alpha}{\longrightarrow} E^V \overset{d\Phi|_{E^V}}{\longrightarrow} E^{\prime V}
\overset{\alpha^{\prime-1}}{\longrightarrow} E'
$$
covering the identity of $M$, thus showing the $E$ and $E'$ are isomorphic as vector bundles.
Here's the less general proof I originally posted.
Suppose first that $\pi\colon E\to M$ is a smooth fiber bundle with $k$-dimensional model fiber $F$. There is a rank-$k$ vector bundle $T^V E\to E$, called the vertical tangent bundle, whose fiber at a point $p\in E$ is the tangent space to the fiber $E_{\pi(p)} = \pi^{-1}(\pi(p))$: in other words, $T^V_pE = T_p(E_{\pi(p)}) = \ker d\pi_p$. If $E$ has a global section $\sigma\colon M\to E$, then $T^V E$ pulls back to a vector bundle over $M$, which I'll denote by $E^V = \sigma^*(T^V E)$ with projection $\pi^V\colon E^V\to M$.
Now suppose $E$ has a global trivialization (as a fiber bundle) $\Phi\colon E\to M\times F$. Thus $\Phi$ is a diffeomorphism satisfying $\pi_1\circ\Phi = \pi$ (where $\pi_1\colon M\times F\to M$ is the projection on the first factor).
Because $\Phi$ is a bundle map, the global differential $d\Phi\colon TE\to T(M\times F)$ restricts to a bundle isomorphism from $T^V E$ to $T^V (M\times F)$, and therefore $T^V E$ is trivial.
It follows that $E^V$ is also trivial, since it's the pullback of a trivial bundle.
Now suppose $E$ also has the structure of a smooth vector bundle.
The zero section is a smooth global section, so we obtain the pullback vertical bundle $E^V$ as before, whose fiber at each point $q\in M$ is $T_0(E_q)$. In this case, since $E_q$ has the structure of a finite-dimensional vector space, the tangent space $T_0(E_q)$ is canonically isomorphic to the vector space $E_q$ itself; the isomorphism is given by sending an element $v\in E_q$ to the derivation $D_v\colon C^\infty(E_q) \to \mathbb R$ defined by $D_v(f) = (d/dt)|_{t=0} f(tv)$. Putting together these isomorphisms for all $q\in M$ shows that the vector bundle $E$ is canonically isomorphic to $E^V$, provided the map $\alpha\colon E\to E^V$ so obtained is smooth.
In a neighborhood $U$ of any point of $M$, there is a local vector bundle trivialization $\Psi\colon \pi^{-1}(U)\to U\times \mathbb R^k$. Its differential restricts to a smoooth local trivialization $d\Psi|_{(\pi^V)^{-1}(U)}\colon (\pi^V)^{-1}(U) \to U\times \mathbb R^k$. Unwinding the definitions shows that the map $d\Psi\circ\alpha\circ \Psi^{-1}\colon U\times \mathbb R^k\to U\times \mathbb R^k$
has the form $d\Psi\circ\alpha\circ \Psi^{-1}(q,v) =(q,v)$.
Since $\Psi$ and $d\Psi|_{(\pi^V)^{-1}(U)}$ are diffeomorphisms, this shows that $\alpha$ is smooth in a neighborhood of each point.
Best Answer
Over spheres, you actually have a nice way to distinguish bundles.
Every sphere $S^n$ can be written as a union of two contractible spaces, $U \cup V$, whose intersection is a small neighborhood of the equator, $\mathbb{R} \times S^{n-1}$. For instance if $x,y$ are the north and south poles, respectively, you can take $U = S^n - \{x\}, V = X^n - \{y\}$.
Over $U$ and $V$, any bundle is trivial, as you noted. The bundle on the sphere is glued together from these two trivial bundles.
If you're interested in a vector bundle, the transition function for the bundle is given by a map $U \cap V \to GL_m$, for a vector bundle of rank $m$. If this map is null-homotopic, meaning that it can be homotoped to the constant map, then your bundle is trivial, as any trivialization over $U$ can then be glued to a trivialization over $V$. So in the end you are reduced to the study of maps from $U \cap V \simeq S^{n-1}$ to your structure group -- i.e., $\pi_{n-1}GL_m$.
For $m=1$, this amounts to asking whether any map $S^{n-1} \to GL_1$ is null-homotopic. In your case for $S^3$, since $GL_1(\mathbb{R})$ is a disjoint union of two contractible spaces, and $S^2$ is connected, any map is null-homotopic.
In the fiber bundle case with fiber $S^1$ and base $S^3$, you are looking at maps $S^2 \to S^1$. But $\pi_2(S^1) = *$ so any circle bundle is contractible over $S^3$.
Note that both facts hold for higher spheres as well.