Connection under the frame change. In the below picture ,I don't know how to compute the red line 2. But if accept the red line 2, I  can compute the red line 3. But there still be a little question, that at last ,I have 
$$
\Gamma_{ib}^aV^b=V^bF_j^a\partial_iF_b^j+...
$$
$$
\nabla_iV^a=V^aF_j^a\partial_iF_a^j+...
$$
The $F_j^a=(F_a^j)^{-1}$, at the right part ,I always have three $a$ , it is not same as above. 
I want to know how to compute the red line 2. So thanks for detail answer or hint.
==================================================================



 A: (1) Fix a coordinate vector field $E_i$ for $g(0)$ For $g(t)$,
consider a frame $$ F_a=F_a^i(t) E_i $$ s.t. $$ g(t)(F_a(t),
F_b(t))=\delta_{ab} \ \ast$$ for any $t$
And if we assume that $$ \partial_t F_a^i(t)= g^{ij} R_{jk} F^k_a \ \ast\ast$$
 so   $$  \frac{\partial }{\partial_t} [ g(t)(F_a(t),
F_b(t)) ] $$ $$= -2Ric(g(t)) (F_a(t),F_b(t)) +g_{im} F_b^m g^{ij}
R_{jk} F^k_a + g_{mt} F^m_a g^{tj} R_{jk} F^k_b =0$$
That is, solution in $\ast\ast$ is the desired one, which satisfies
$\ast$.
(2) Define $$ h:= F^\ast g,\ h_{ij}:= g_{\alpha\beta} F_i^\alpha
F^\beta_j $$
That is by (3), $$ \Gamma(h)_{ij}^k = \Gamma(g)_{\alpha\gamma}^\tau
 F^\gamma_jF^\alpha_i F_\tau^k + E_i(F^\alpha_{j})
F_\alpha^k$$ $$= \Gamma(g)_{i\gamma}^\tau
 F^\gamma_j F_\tau^k + E_i(F^\alpha_{j})
F_\alpha^k$$
So if $V=V^\alpha F_\alpha$, $$ \nabla_i V^\gamma= E_i V^\gamma +
\Gamma_{i\alpha}^\gamma V^\alpha $$
In further $$ \nabla_\beta V^\gamma =F_\beta^i \nabla_i V^\gamma  $$
(3) Assume that $ f(x)=y$ is a
diffeomorphism and $$h:=f^\ast g
$$ Define
$$ g_{\alpha\beta }:= g(\partial_{y_\alpha},\partial_{y_\beta}) $$
$$  h_{ij}:= h(\partial_{x_i},\partial_{x_j}) $$
$$ f^\alpha_i:= \frac{\partial y^\alpha}{\partial x_i  } $$
So $$ \Gamma(h)_{ij}^k = \Gamma(g)_{\alpha\gamma}^\tau
 f^\gamma_jf^\alpha_i f_\tau^k + f^\alpha_{ij}
f_\alpha^k$$
