Why is $\frac{d}{d \mu} \big|_{\mu=0} \, F^*_\mu F^*_\lambda t = F^*_\lambda \frac{d}{d \mu} \big|_{\mu=0} \, F^*_\mu t $? In the book "Manifolds, Tensor Analysis, and Applications" by Marsden, Ratiu, Abraham the following relation (see the proof of 6.4.1, third edition) is used:
$$\frac{d}{d \mu} \bigg|_{\mu=0} \, F^*_\mu F^*_\lambda t = F^*_\lambda \frac{d}{d \mu} \bigg|_{\mu=0} \, F^*_\mu t $$
Here $t$ is an arbitrary tensor field $t \in \mathcal T^r_s$, $F$ being a flow, $F^*$ its pullback. However I don't see why this is true. Particularly I'm interested in the simplest case of t being just a smooth function ($t \in \mathcal T^0_0$).

Result : If $\phi$ and $\psi$ are $1$-parameter groups s.t. $$ \psi'= X$$
 then $$\phi^\ast (L_X\alpha) = \phi^\ast \lim_t \frac{\psi(t)^\ast
 \alpha - \alpha }{t} = \lim \frac{(\phi^{-1} \circ \psi \circ \phi
 )^\ast
 (\phi^\ast \alpha )-\phi^\ast \alpha }{t} $$
Here recall $$ \frac{d}{ds}\bigg|_{s=0}
 \phi^{-1}\circ \psi (s,\phi (z)) =d\phi^{-1} X_{\phi(z)} = \phi^\ast X $$ so that $$
 \phi^\ast (L_X\alpha) = L_{\phi^\ast X} \phi^\ast\alpha $$
Returning to OP : Let $X:= \frac{\partial }{\partial \mu}\bigg|_{\mu =0
  }F_\mu$. Then $$ \frac{d}{d\mu}\bigg|_{\mu=0} F_\mu^\ast
  F^\ast_\lambda t=\frac{d}{d\mu}\bigg|_{\mu=0} F_\lambda^\ast
  F^\ast_\mu t $$
and $$ F_\lambda^\ast
 \frac{d}{d\mu}\bigg|_{\mu=0} F_\mu^\ast
 t= F^\ast_\lambda L_X t=L_X   F^\ast_\lambda t=\frac{d}{d\mu}\bigg|_{\mu=0} F_\mu^\ast
  F^\ast_\lambda t $$
