Extending a vector to a section of a vector bundle parallel with respect to a given connection 
Let $E$ be a vector bundle on a smooth manifold $M$, let $\nabla$ be a connection on $E$. Let $v\in E_p$, where $p\in M$. Is true that there exist $s\in\Gamma(M,E)$ such that $s(p)=v$ and $\nabla(s)=0$ ?

 A: For a counterexample consider the Riemannian submanifold $S^2\subseteq\mathbb R^3$ with Levi-Civita connection $\nabla$. Using that $\nabla$ is a metric connection yields that for a parallel tangent vectorfield   $s\in\Gamma(S^2,TS^2)$ and  all tangent vectors $X$
$$X\langle s,s\rangle=\langle\nabla_Xs,s\rangle+\langle s,\nabla_X s\rangle=0$$
which implies that $\langle s,s\rangle$ is a constant function on $S^2$. In particular if $s$ is nonzero at some point it is also nonzero at all the other points. Now by the Hairy ball theorem every continous  tangent vectorfield on $S^2$ has a zero, so if you start with $s(p)\neq 0$ there does not exist a parallel extension.
A: It's not possible in general. For example, suppose $M$ is a Riemannian manifold, $E=TM$, and $\nabla$ is the Levi-Civita connection. If it were always possible to find such a section, you could begin with an orthonormal basis for $T_pM$ and extend it to a parallel orthonormal frame. This is impossible unless the Riemannian metric is flat.
A: In fact the statement is far from being true: Generically, given a vector bundle $E \to M$ equipped with a vector bundle connection $\nabla$ and a choice of point $p \in M$, the only vector $v \in E_p$ that admits an extension to a parallel section $s \in \Gamma(E\vert_U)$ to any neighborhood $U \ni p$ is the zero vector.
Indeed, it follows from the definition of the curvature $$R \in \textstyle{\Gamma\left(\bigwedge^2 T^*M \otimes \operatorname{End}(E)\right)}$$ of $\nabla$ that it annihilates any parallel (local) section $s$ of $E$, that is, that $$R(\,\cdot\,,\,\cdot\,)(s) = 0$$---that is, that $R$ is degenerate.
Example For any (local) vector field $X$ on $S^n$, the curvature $R$ of the Levi-Civita connection of the round metric satisfies $$R(\,\cdot\,,\,\cdot\,)(X) = X^\flat \wedge \operatorname{id} .$$ In particular (for $n > 1$), if $\nabla X = 0$ then $X = 0$.
It's maybe worth mentioning that the statement is true in the special case that $\dim M = 1$, at least locally (and globally if $M$ is simply connected). After all, if $M$ is $1$-dimensional, there is essentially only one path along which a vector can be parallel-transported.
