Let us consider some vector bundles over $S^4$. Since $S^4$ is simply connected all vector bundles are oriented and rank $k$ vector bundles over $S^4$ are classified by $\pi_{3}(SO(k))$ by the clutching construction. Now $\pi_3(SO(1))$ and $ \pi_3(SO(2)=\pi_3(S^1)$ are trivial groups, so there do not exist non-trivial rank $1$ and rank $2$ bundles over $S^4$. What about rank $3$? Well $SO(3)\cong \mathbb RP^3$ and the long exact sequence of the fiber bundle $$ \mathbb{Z}_2\rightarrow S^3\rightarrow \mathbb{RP}^3 $$ shows that $\pi_3(SO(3))=\mathbb{Z}$. Hence there are $\mathbb{Z}$ different rank $3$ vector bundles over $S^4$.

Only the trivial one admits a non-zero section. If another one admits a section, then the orthogonal complement (a rank $2$ bundle) must be non-trivial. But our previous computation shows that these do not exist.

No.

Let us consider some vector bundles over $S^4$. Since $S^4$ is simply connected all vector bundles are oriented and rank $k$ vector bundles over $S^4$ are classified by $\pi_{3}(SO(k))$ by the clutching construction (I am glossing over basepoint issues here, but I think it is correct in this setting). Now $\pi_3(SO(1))$ and $ \pi_3(SO(2)=\pi_3(S^1)$ are trivial groups, so there do not exist non-trivial rank $1$ and rank $2$ bundles over $S^4$. What about rank $3$? Well $SO(3)\cong \mathbb RP^3$ and the long exact sequence of homotopy groups of the fiber bundle $$ \mathbb{Z}_2\rightarrow S^3\rightarrow \mathbb{RP}^3 $$ shows that $\pi_3(SO(3))\cong\pi_3(\mathbb{RP}^3)\cong \mathbb{Z}$. Hence there are $\mathbb{Z}$ different rank $3$ vector bundles over $S^4$.

Only the trivial one admits a non-zero section. If another one admits a section, then the orthogonal complement (a rank $2$ bundle) must be non-trivial, otherwise the bundle is trivial. But our previous computation shows that there are no-nontrivial rank $2$ bundles over $S^4$.