We supposed $M$ is a compact ( i.e, closed without boundary) manifold. Let $\phi_{t}: M \mapsto M$ be a one-parameter family of flow and $p: \widetilde{M} \mapsto M$ be the universal cover of $M$. Let $X$ be the vector field associated to the flow $\phi_{t}$ on $M$. Since the universal covering map is a local diffeomorphism, we can take the pullback of the associated vector field to the universal cover, which will give us a new flow $\widetilde{\phi_{t}}$ on $\widetilde{M}$. We call it a lift of the flow.
I wanted to see why can we extend this lifted flow for all time $t$ in the universal cover, as it's no longer compact. Also is it true that the flow will commute with the monodromy action of $\pi_{1}(M)$ on $\widetilde{M}$. That is to say for $\tilde{x}$ $\in$ $\widetilde{M}$, is it true that
$\gamma . \widetilde{\phi_{t}}(\tilde{x})= \widetilde{\phi_{t}}(\gamma . \tilde{x}) $, where $\gamma$ is an element $\pi_{1}(M)$. Thanks in advance.
The reason we can extend the flow upstairs for all time is that every flow line downstairs in $M$ can be lifted upstairs to $\widetilde M$.
To be precise, take any $\tilde x \in \widetilde M$, and let $x = p(\tilde x)$. As you say, by compactness the flow line $\phi_t(x)$ is well defined in $M$ for all $t \in \mathbb{R}$. The function $$f_x : \mathbb{R} \mapsto M $$ defined by $$f_x(t) = \phi_t(x) $$ is obviously continuous. By the lifting lemma from covering space theory, and using that the domain $\mathbb{R}$ of $f_x$ is simply connected, there exists a unique function $$F_{\tilde x} : \mathbb{R} \mapsto \widetilde M $$ such that $F_{\tilde x}(0)=\tilde x$ and such that $p \circ F_{\tilde x} = f_x$. Now if you trace through the definitions, you will see that $F_{\tilde x}$ is a flow line, meaning that $$\frac{d F_{\tilde x}}{dt} \mid_{t=t_0} = X(F_\tilde x(t_0)) $$
For your second question, the equality of those two sides is true because both sides represent flow lines of the lifted vector field $\widetilde X$ based at $\gamma \cdot \tilde x$ (as you can check), but the flow line based at $\gamma \cdot \tilde x$ is unique, by the existence and uniqueness theorem for ODEs. I presume that your vector field $X$ is regular enough so that the existence and uniqueness theorem can be applied. Smoothness of $X$ implies smoothness of $\widetilde X$ which is more than enough regularity.