Geodesics of The Tangent Bundle

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?

Yes, your conjecture is true. To prove it, we should calculate the geodesic equation for the tangent bundle. For a curve $$t \mapsto (x(t), y(t)) \in TM$$ such that $$x(t)$$ is not a fixed point, it is a geodesic if and only if $$\nabla_{\dot x} \dot x = -R (y, \nabla_{\dot x} y) \dot x, \quad \nabla_{\dot x} \nabla_{\dot x} y = 0.$$
This formula is given as Corollary 4.4 in Geodesics on tangent bundles with horizontal Sasaki gradient metric by Abderrahim Zagane, but the condition $$\dot x \neq 0$$ is missing. Rigorously, you can deduce it from general Levi-Civita connection formulae given in Curvature of the Induced Riemannian Metric on the Tangent Bundle of a Riemannian Manifold by Oldrich Kowalski. In the case when $$x(t)$$ is a fixed point, the geodeic equation becomes $$y^{\prime\prime}(t) = 0$$.
In your conjecture, we have $$\nabla_{\dot x} y = 0$$ by definition of parellel transport and $$\nabla_{\dot x} \dot x = 0$$ since $$x$$ is a geodesic on $$M$$.