$\newcommand{\al}{\alpha}$ Let $M$ be a smooth Riemannian manifold. Fix $p \in M, v,w \in T_pM$, and define $\gamma(t) = \exp_p(t(v + \frac{t}{2}w))$.
Is there a coordinate-free proof that $(\nabla_{\frac{\partial}{\partial t}}\dot \gamma)(0)=w$? Even without using normal coordinates?
An attempt using Jacobi Fields:
Define $f(s,t)=\exp_p(t(v+sw))$. Then $\gamma(t)=f(t/2,t)$, so $$ \dot \gamma(t)=1/2\frac{\partial f}{\partial s}(t/2,t)+\frac{\partial f}{\partial t}(t/2,t). $$ Thus (If I am applying the chain rule correctly?) $$ \nabla_{\frac{\partial}{\partial t}}\dot \gamma (0)=1/2\big(1/2\frac{D}{\partial s}\frac{\partial f}{\partial s}(0,0)+\frac{D}{\partial t}\frac{\partial f}{\partial s}(0,0)\big)+1/2\frac{D}{\partial s}\frac{\partial f}{\partial t}(0,0)+\frac{D}{\partial t}\frac{\partial f}{\partial t}(0,0)= $$ $$ 1/2\big(1/2\frac{D}{\partial s}\frac{\partial f}{\partial s}(0,0)+\dot J(0)\big)+1/2\frac{D}{\partial s}\frac{\partial f}{\partial t}(0,0), $$ where $J(t)=\frac{\partial f}{\partial s}(t,0)$ is the Jacobi field associated with the geodesic variation $f(s,t)$. We have $\dot J(0)=w$.
Thus, so far we obtained $$\nabla_{\frac{\partial}{\partial t}}\dot \gamma (0)= w/2+1/4 \frac{D}{\partial s}\frac{\partial f}{\partial s}(0,0)+1/2\frac{D}{\partial s}\frac{\partial f}{\partial t}(0,0). $$ I am not sure how to proceed.
When using normal coordinates, this is easy: In these coordinates $\gamma(t)=tv+\frac{t^2}{2}w$, so $$ \dot \gamma(t)=v+tw=(v^i+tw^i)\partial_i(\gamma(t)). $$ Thus $$ (\nabla_{\frac{\partial}{\partial t}}\dot \gamma)(0)=w^i\partial_i(p)+v^i\nabla_{\frac{\partial}{\partial t}}\partial_i(\gamma(t))=w^i\partial_i(p)=w, $$ where we used the fact that $$\big( \nabla_{\dot \gamma(t)}\partial_i\big)(p)=0, $$ which follows from the fact that the Christoffel symbols of normal coordinates vanish at the origin. (after expanding $\dot \gamma(t)$ in terms of the coordinate vector fields).