Let $A$, $B$ be smooth submanifolds of a smooth manifold $M$ and $X\in C^\infty(TM)$ a vector field such following its flow $\xi^X$ gives a diffeomorphism between $A$ and $B$. Suppose also that $P\subseteq C^\infty(TM)$ is a finite dimensional subspace such that, for every $E\in P$, following $\xi^{X+E}$ also gives a diffeomorphism between $A$ and $B$ (in my particular setup $A$ and $B$ are regular levels of a smooth function with no critical levels in between, $X$ is a gradient-like vector field and the elements of $P$ are used to perturb $X$).
Consider the map $$ \begin{array}{rccc} \psi:&A\times P&\longrightarrow & B\\ &(p,E)&\longmapsto &\xi^{X+E}_{t_{p,E}}(p) \end{array}. $$ where $t_{p,E}$ is the exact time needed to reach $B$ from $p$ under de flow of $X+E$.
Consider also the map $$ \begin{array}{rccc} \Phi_{p,t}:&P&\longrightarrow & M\\ &E&\longmapsto &\xi^{X+E}_t(p) \end{array}. $$
Fixing $x\in A$ we have $\psi_x:=\psi(x,\cdot):P\rightarrow B$.
What is the relation between $d_E\psi_x$ and $d_E\Phi_{x,t_{x,E}}$?
I'm going to assume your notation means that the $E$ in the (sub-)subscript of $\Phi_{x, t_{x,E}}$ is not varying with your derivative (otherwise it's pretty straightforward to check that your two derivatives are equal).
For each $x$ you have a map $\tau_x : P \to \mathbb{R}$ given by $\tau_x(E) = t_{x,E}$. With these notations we'll show $$ d\psi_x|_E = d\Phi_{x, \tau_x(E)}|_E + (X + E)_{\psi(x,E)} \cdot d\tau_x|_E, $$ where the rightmost term is to be interpreted in the obvious way as a map $T_E P \to T_{\psi(x,E)}B$.
Checking this is just a matter of applying definitions and the chain rule. We have a map $\xi : P \times \mathbb{R} \to M$ with $\xi(E, s) = \xi^{X+E}_s$. We use the notations \begin{align} (\partial_P \xi)|_{E, s} &: T_E P \to T_{\xi(E, s)} M \\ (\partial_{\mathbb{R}} \xi)|_{E, s} &: \mathbb{R} \to T_{\xi(E,s)} M \end{align} for the partial derivatives. We can compute these more explicitly: for the $P$ partial, we have $$ (\partial_P\xi)|_{E, s} = d\Phi_{x, s}|_E $$ while for the $\mathbb{R}$ partial, $$ (\partial_{\mathbb{R}}\xi)|_{E, s} = (X + E)_{\psi(x, E)}. $$ (These both follow immediately from the definitions.)
Then given $E, F \in P$, the element $d\psi_x|_E(F)$ of $T_{\psi(x,E)}B$ is given by \begin{align} d\psi_x|_E(F) &= \frac{d}{dt}\Big|_{t = 0} \xi\big(E + tF, \tau_x(E + tF)\big) \\ &= (\partial_P \xi)|_{E, \tau_x(E)}(F) + (\partial_{\mathbb{R}}\xi)|_{E, \tau_x(E)} \cdot d\tau_x|_E(F) \\ &= d\Phi_{x, \tau_x(E)}|_E(F) + (X + E)_{\psi(x, E)} \cdot d\tau_x|_E(F) , \end{align} as claimed.