Suppose $(M,g)$ is a complete Riemannian manifold. We can endow the tangent bundle $TM$ with a natural choice of metric, the so called Sasaki metric given by:
$\alpha(t)=(p(t),v(t))$ , $\beta(t)=(q(t),w(t))$. Then define $\langle\alpha^\prime(0),\beta^\prime(0)\rangle=\langle p^\prime(0),q^\prime(0)\rangle+\langle\frac{Dv}{dt}(0),\frac{Dw}{dt}(0)\rangle$. In other words we identify the vertical vectors at $T_{(p,v)}TM$ with $T_pM$ and the Euclidean metric given on that by $g$. Then we pullback $g$ by $D\pi: H \to TM$ to the horizontal vectors and define $H$ and $VE$ to be orthogonal to each other.
In this metric one can prove if $\gamma(t)$ is a geodesic and $v(t)$ a parallel vector field along $\gamma$, then $(\gamma(t),v(t))$ is a geodesic in $TM$. Now suppose $(p,v)$ and $(q,w)$ are two arbitrary points in $TM$ and $\gamma$ is a geodesic with $\gamma(0)=p, \gamma(1)=q$. Consider $v(t),w(t)$ which are the parallel transports of $v$ and $w$ along $\gamma$.
Is it true that the curve $\delta(t)=(\gamma(t),(1-t)v(t)+tw(t))$ is a geodesic between $(p,v)$ and $(q,w)$? Notice that if $w$ is the parallel transport of $v$ at $t=1$ then $\delta(t)$ is a geodesic.
In other words this is equivalent to say if $\delta(t)=(\gamma(t),v(t))$ is a curve in $TM$ such that $\gamma$ is a geodesic in $M$ and $\frac{D^2v}{dt^2}=0$, then $\delta$ is a geodesic in $TM$.
Do you have any counterexample or proof?