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.
Every smooth vector bundle $E$ admits global sections. For example, let $U \subset M$ be a trivialising open set, then for every $\sigma \in \Gamma(U, E|_U)$ and every function $\rho : M \to \mathbb{R}$ with support contained in $U$, then $\rho\sigma$ is a global section of $E$, i.e. $\rho\sigma \in \Gamma(M, E)$.
If $\Phi : E \to F$ is a vector bundle homomorphism, then there is an associated section $\sigma : X \to \operatorname{Hom}(E, F)$ of $\pi$ given by $x \mapsto \Phi_x$. Conversely, If $\sigma : X \to \operatorname{Hom}(E, F)$ is a section of $\pi$, then we obtain a vector bundle homomorphism $\Phi : E \to F$ given by $e \mapsto \sigma(\pi(e))(e)$; that is, if $e \in E_x$, we have $e \mapsto \sigma(x)(e)$.
Best Answer
Since $H^1(S^2, \mathbb{Z}_2) = 0$, any real line bundle over $S^2$ is trivial and thus admits a nowhere-vanishing section. But $TS^2$ does not (because, e.g., $e(TS^2) = \chi(TS^2)$ is nonzero), so $TS^2$ has no line subbundle.