Let $f=f(t)$ be a physical quantity of a system, $t$ is a variable e.g time. For infinitesimal variation $\delta t$: $$f(t_{0}+\delta t)=f(t_{0})+\frac{\mathrm{d}f}{\mathrm{d}t}\delta t=f_{0}+\{f,H_{T}\}$$ where $H_{T}=H+u_{m}\phi_{m}$ with $\phi_{m}\approx 0$ is primary constraint of the system. Afterward, $f(t_{0}+\delta t)$ can be written as: $$f(t+\delta t)=f_{0}+\{f,H\}\delta t + u_{m}\{f,\phi_{m}\}\delta t$$ This is the way we apply gauge transformation with the infinitesimal change of $t$.
I think, if we apply the gauge transformation one more time, it will look like: $$f'=f_{0}+\{f(t_{0}+\delta t),H\}\delta t + u_{m}\{f(t_{0}+\delta t),\phi_{m}\}\delta t$$ Then the procedure will be continued with expansion of $f(t_{0}+\delta t)$. Am I correct? Assume that I am correct at that point, the calculation continues, we obtain: $$f'=f_{0}+\{f_{0},H\}\delta t+\{\{f,H\},H\}\delta^2 t+\{u_{n}\{f,\phi_{n}\},H\}\delta^2 t+u_{m}\{f_{0},\phi_{m}\}\delta t+u_{m}\{\{f,H\},\phi_{m}\}\delta^2 t+u_{m}\{u_{n}\{f_,\phi_{n}\},\phi_{m}\}\delta^2 t=f_{0}+\{f_{0},H\}\delta t+u_{m}\{f_{0},\phi_{m}\}\delta t $$ I smell something wrong here because after applying second gauge, I can not see where its effect is. There is no appearance of $\phi_{n}$ and $u_{n}$. Finally, how does it look if we apply $n$ gauge transformations? For related documents:
http://homepages.uni-r.de/~bon39708/lectures/2017_ws/skript_ws_2017.pdf page 18
http://crypto.fmf.ktu.lt/lt/telekonf/archyvas/KriptoTeorija/2001%20-%20Paul%20Adrien%20Maurice%20Dirac%20-%20Lectures%20on%20Quantum%20Mechanics.pdf page 20-21 of Dirac's lectures.
Update with answer of my question
I had just figured it out how to apply two successive gauge transformation.
At first, calculate $\Delta f=\epsilon_{m}\{f,\phi_{m}\}$, then first gauge is $f=f_{0}+\Delta f=f_{0}+\epsilon_{m}\{f,\phi_{m}\}$.
Next, treat $f+\epsilon_{m}\{f,\phi_{m}\}$ as new $f$, we have $$f'=f_{0}+\epsilon_{m}\{f_{0},\phi_{m}\}+\delta t\{f+\epsilon_{m}\{f,\phi_{m}\},H\}+u_{n}\{f+\epsilon_{m}\{f,\phi_{m}\},\phi_{n}\}\delta t$$compute $\Delta f'$, we get $\epsilon_{n}\{f+\epsilon_{m}\{f,\phi_{m}\},\phi_{n}\}$. Thus, after appying two successive gauge, what we get is $$f=f_{0}+\epsilon_{m}\{f,\phi_{m}\}+\epsilon_{n}\{f+\epsilon_{m}\{f,\phi_{m}\},\phi_{n}\}$$
That's how it is done. Still, I wonder why $\epsilon_{m}\phi_{m}$ is generator of gauge.