Let $(M, \omega)$ be a symplectic manifold on which the Lie group $G$ acts symplectically via $\phi$. That is $$G \times M \rightarrow M,$$ $$(g,m)\rightarrow gm=\phi_g (m)$$ where all the $\phi_g$ $g \in G$ are diffeomorphisms. Also let $\mathfrak{g}$ and $\mathfrak{g}^*$ be respectively the Lie algebra of $G$ and its dual.
For $X \in G$, denote by $X_M$ the vector field on $M$ which for all $m \in M$ is given by the rule $$X_Mf(m):=\frac{d}{dt}f(\phi_{\textrm{exp }tX}(m))\bigg |_{t=0}$$ for $f$ being a function on $M$.
$\textbf{My question(s)}:$ I've read elsewhere that if we have a one-parameter family of diffeomorphisms, say $\phi^{'}_{t}$, and a smooth function $f$, then $f(\phi^{'}_{t}(a))$ is a smooth function of $t$ and $$\frac{\partial}{\partial t}f(\phi^{'}_{t}(a))\bigg |_{t=0} = X_a f$$. I'm trying to relate this to the above statements. What is this really saying? Why does evaluating $f$ at the image of this diffeomorphism and taking its differential yield the same result as applying a vector field to $f$. Also, $\textbf{most importantly}$, what does the symbol $\phi_{\textrm{exp }tX}$ actually mean? Is that an integral curve?
Most of this comes from $\textit{Introduction to Symplectic Geometry}$ by Rolf Berndt and I've also looked at the relevant section of the text by McDuff and Salamon but any clarification is much appreciated because I couldn't get it from these works or Da Silva's notes or even from other answers to questions related to this topic.