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.
For the equivalence of these definitions, I would look here: Local triviality of principal bundles.
The existence of a $G$-equivariant cover is equivalent to the existence of $G$-valued transition functions:
Suppose $(U_\alpha,\Phi_\alpha)$, $\Phi_\alpha : P\vert_{U_\alpha} \to U_\alpha\times F$, is a trivializing cover. This defines a collection of maps $\phi_\alpha : P\to F$ by
$$
\Phi_\alpha(p) = (\pi(p), \phi_\alpha(p)).
$$
For a right principal $G$-bundle, this covering is $G$-equivariant if $\phi_\alpha(pg) = \phi_\alpha(p)g$. Now we have
$$
\Phi_\alpha \circ \Phi_\beta^{-1} : U_\alpha \cap U_\beta \times F \to U_\alpha \cap U_\beta \times F
$$
is an isomorphism of trivial $G$-bundles and so takes the form
$$
(x, f) \mapsto (x, h_{\alpha\beta}(x,f)).
$$
If the covering is $G$-equivariant then so is this map, which means that $h_{\alpha\beta}(x,fg) = h_{\alpha\beta}(x,f)g$. Since $G$ is acting freely and transitively, fixing a point of $F$ identities $F$ with $G$ and $h_{\alpha\beta}$ is entirely determined by the function $g_{\alpha\beta}: U_\alpha\cap U_\beta \to G, x \mapsto h_{\alpha\beta}(x,e)$. Thus the transition functions are given by left-multiplication by $g_{\alpha\beta}$. This is what is meant by the transition functions being $G$-valued.
Conversely, if the transition functions are $G$-valued then the trivializations will be $G$-equivariant. This is because
$$
P = \sqcup_\alpha U_\alpha \times F/\sim, ~~ (x, f) \sim (x, g_{\alpha\beta}(x)f) \text{ for } x \in U_\alpha\cap U_\beta.
$$
The equivariance then comes from the fact that the transition functions are operating by left-multiplication, while the $G$-action is right multiplication.
Best Answer
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?