Let $X \in \Gamma(M)$ with $\Phi^{X}_{t}(p)$ as the one parameter group at a point $p \in M$. I believe that in general the equality $$T_p\Phi^{X}_{t_0}(X_p)=X_{\Phi^{X}_{t_0}(p)}=(\frac{\partial}{\partial t}\Phi^{X}_{t}(p))(t_0)$$ (i.e. $(\Phi^{X}_{t_0})_*(X_p)=X_{\Phi^{X}_{t_0}(p)}$) does not hold. I'm looking for a simple counterexample. Thanks in advanced.
Adendum: I want to give a bit of context to the question. While studying the Lie derivative, I wondered whether the vector field (which generates the flow) would be invariant under the pushforward of the one parameter group.
As shown below this is not the case. However I do wonder under which conditions this should indeed hold. Say, for complete flows.
EDIT: Yes, your formula is correct. What's going on here is the computation of the Lie derivative $\mathscr L_XX$, which is always $0$. Of course, for a general vector field $Y$, $\mathscr L_XY$ will no longer be $0$.
My original counterexample is now a non-counterexample:
Take $X = x^2\dfrac{\partial}{\partial x}$ on $\Bbb R-\{0\}$. Then $\Phi_t(x) = \dfrac x{1-tx}$ for $x\ne 0$ and for $|t|<1/|x|$. Then $\dfrac d{dx}\Phi_t(x) = \dfrac 1{(1-tx)^2}$. In your notation, $T_p\Phi_t = \dfrac1{(1-tp)^2}$. Applying this to $X_p = p^2\dfrac{\partial}{\partial x}$, we get $\dfrac1{(1-tp)^2}\cdot p^2 \dfrac{\partial}{\partial x} = \left(\dfrac p{1-tp}\right)^2 \dfrac{\partial}{\partial x} = X_{\Phi_t(p)}$, as predicted.
Here now is a proof of your formula: If we consider the curve $$\alpha(t) = \Phi_{t+t_0}(p),$$ then $\alpha'(0) = X_{\Phi_{t_0}(p)}$. On the other hand, setting $\beta(t) = \Phi_t(p)$, then of course $\alpha(t) = \Phi_{t_0}\circ\beta(t)$, so by the chain rule $$\alpha'(0) = T_p\Phi_{t_0}\beta'(0) = T_p\Phi_{t_0}X_p.$$ This is your equality.
(By the way, this hideous $T_p$ notation, rather than $Df$ or $f_*$, is very annoying. :))