We denote the flow, of the dynamical system defined on $M$ by $$ \dot x(t) = X(x(t)), $$ by $\exp(tX) ~x(0)$, which is a map from $M$ to $M$.
This notation comes from the linear case, where the flow is the matrix exponential.
For two vector fields $X,Y$, the Baker-Campbell-Hausdorff formula reads $$ \exp(sX)\exp(tY)=\exp(\zeta(X,Y)) ~, $$ where $\zeta(X,Y)$ belongs to the Lie Algebra generated by $X$ and $Y$, with $$ \zeta(X,Y)=sX + tY + st/2[X,Y]+\cdots $$
But what is the definition of $\exp(\zeta(X,Y))$ ?
Abelian nonlinear Lie advection, ubiquitous in the RG in physics, evolves a variable $x(0)=z$ possibly nonlinearly, to $x(t)=f(t,z)=\exp (tX(z)) z$. Operator exponentials are defined by the series expansion thereof w.r.t. the operators, and the CBH expansion is but the combinatorics of two operators involved in sequential expansion.
So, as you indicated, $$ e^{\zeta (X,Y) }= e^{sX + tY + st[X,Y]/2+\cdots}, $$ even for fiercely nonlinear operators on z.
Their action which specifies the resulting trajectory is simply $$ e^{\zeta (X,Y) } z= z+\zeta (X,Y) z + \zeta^2 z/2+...= z+ (sX + tY + st[X,Y]/2+\cdots)z + ... $$
You might illustrate the point by choosing $$ X= z\frac{d}{dz} \qquad Y=z^2 \frac{d}{dz}~~, $$ and evaluating their simple commutators, $[X,Y]=Y$, so, then, $\zeta=sX+stY/(1-\exp(-s))$. This is one of those felicitous circumstances where you can effectively sum the CBH expansion completely. In practice, of course, you use logic, instead.
You thus have $$ e^{sz \partial_z } e^{tz^2 \partial_z } f(z)= e^{sz \partial_z } f\left (\frac{z}{1-tz}\right )=f\left (\frac{e^s z}{1-te^s z}\right )= f\left (\frac{ z}{e^{-s}-t z}\right ) . $$ Looking at $f(z)=z$, you appreciate the trajectory $$ z\mapsto z+sz+tz^2+2stz^2+... $$