Let $u_t$ be a time dependant vector field on a Riemannian mfd $M$ with integral curve $\gamma_t$ such that $\gamma'_t=u_t(\gamma_t)$ and $\gamma_0=a\in M$.
It is true that$\frac{D}{dt}(u_t(\gamma_t))=\partial_tu+\nabla_{u_t}u_t$ (along $\gamma$)?
This is my computation: choosing a chart $(x_i)$ write $u_t(x)=u^i(x)\partial_i$ and $\gamma_t=\gamma^i_tx_i$. Then
$\frac{D}{dt}(u_t\circ\gamma_t)=(u^i_t\circ\gamma_t)'\partial_i+u^i_t\circ\gamma_t\nabla_{\gamma_t'}\partial_i=(\partial_tu^i_t\circ\gamma_t+Du^i_t(\gamma_t)[\gamma'_t])\partial_i+u^i_t\circ\gamma_t(\gamma_t^j)'\Gamma_{ij}^k\partial_k=$
(for simplicity of notation I stop writing $\circ\gamma_t$)
$=\partial_tu_t+(u_t^k\partial_ku^i_t)\partial_i+u^i_tu^j_t\Gamma_{ij}^k\partial_k$
which, by changing $i$ with $k$ in the second term, is equal to what I claimed.