Notation Let $G$ be a Lie group and $\pi\colon P\to M$ be a principal $G$-bundle with principal action denoted by $$\Phi\colon G\times P\to P,\ (g,p)\mapsto \Phi(g,p)\equiv \Phi_g(p)\equiv g.p$$
Let us also consider a $G$-invariant vector field $X$ on $P$, i.e. a vector field $X\in\mathfrak{X}(P)$ such that $$(\Phi_g)_\ast X=X, \text{for all}\ g\in G.\tag{$\star$}$$
As you may already know (cf, e.g., Proposition 9.6 in Lee), this condition is equivalent to requiring that, for any $g\in G$ and $t\in\mathbb{R}$, the following identity holds where both sides are defined
$$\text{Fl}^X_t\circ\Phi_g=\Phi_g\circ\text{Fl}^X_t.\tag{$\star\star$}$$
Above, as in the following, we denote the flow of $X$ at time $t$ as $\text{Fl}_t^X\colon\mathcal{U}_t\subseteq P\to P$, while instead, we denote the flow of $X$ as
$$\text{Fl}\colon\cup_{t\in\mathbb{R}}(\{t\}\times\mathcal{U}_t)\subseteq\mathbb{R}\times P\to P, (t,p)\mapsto\text{Fl}^X(t,p)=\text{Fl}^X_t(p).$$
Answer Now we address your question. For simplicity, the answer is split into three smaller results.
Lemma 1 For any $x\in M$, there exist a positive real number $\epsilon_x\in\mathbb{R}_+$ and an open neighborhood $V_x$ of $x$ in $M$ such that, each maximal integral curve of $X$ starting from $\pi^{-1}(V_x)$ is defined at least over $[-\epsilon_x,+\epsilon_x]$, i.e.
$$[-\epsilon_x,+\epsilon_x]\times \pi^{-1}(V_x)\subset\text{dom}(\text{Fl}^X).$$
Proof Pick an arbitrary $p\in\pi^{-1}(x)$. Then there exist a positive real number $\epsilon_p\in\mathbb{R}$ and an open neighborhood $U$ of $p$ in $P$ such that each maximal integral curve of $X$ starting from $U$ is defined at least over the interval $[-\epsilon_p,+\epsilon_p]$, i.e.
$$[-\epsilon_p,+\epsilon_p]\times U\subseteq\text{dom}(\text{Fl}^X).$$
Condition ($\star\star$) assures that, for any $q\in U$, if $\gamma\colon[-\epsilon_p,+\epsilon_p]\to P$ is an integral curve of $X$ with $\gamma(0)=q$, then $\Phi_g\circ\gamma\colon[-\epsilon_p,+\epsilon_p]\to P,\ t\mapsto g.\gamma(t)$, is an integral curve of $X$ with $(\Phi_g\circ\gamma)(0)=g.q$.
Consequently, if one denotes by $V$ the open neighborhood of $x$ in $M$ given by $V=\pi(U)$, one gets that:
- $\pi^{-1}(V)=G.U=\cup_{g\in G}(g.U)$ is a $G$-saturated open neighborhood of $G.p$ in $P$ and
- each maximal integral curve of $X$ starting from $\pi^{-1}(V)$ is defined at least over the interval $[-\epsilon_p,+\epsilon_p]$, i.e.
$$[-\epsilon_p,\epsilon_p]\times\pi^{-1}(V)\subset\text{dom}(\text{Fl}^X),$$
as one needed to prove.
Lemma 2 If $M$ is compact, there exists $\epsilon\in\mathbb{R}_+$ such that each maximal integral curve of $X$ is defined at least on the interval $[-\epsilon,+\epsilon]$, i.e.
$$[-\epsilon,+\epsilon]\times P\subset\text{dom}(\text{Fl}^X).$$
Proof Since $M$ is compact and $\{V_x\}_{x\in M}$ is an open covering of $M$, there exists $x_1,\ldots,x_N\in M$ such that $V_{x_1}\cup\ldots\cup V_{x_N}=M$. Hence, setting $\epsilon:=\min\{\epsilon_{x_1}\ldots,\epsilon_{x_N}\}\in\mathbb{R}_+$, one gets that $[-\epsilon,\epsilon]\times P\subseteq \text{dom}(\text{Fl}^X)$.
Now, one can finally answer your question positively.
Proposition If $M$ is compact, then $X$ is complete.
Proof In view of Lemma 2, for any $p\in P$, each $\gamma\colon[-R,R]\to P$, integral curve of $X$ with $\gamma(0)=p$, can be extended to an integral curve of $X$ defined on $[-R-\epsilon,R+\epsilon]$ defined as follows
$$\tilde\gamma\colon[-R-\epsilon,R+\epsilon]\to P,\ t\mapsto\tilde\gamma(t):=\begin{cases}\text{Fl}_{t-R}^X(\gamma(R)),&\text{if}\ t\in[R-\epsilon,R+\epsilon];\\ \gamma(t), &\text{if}\ t\in[-R,+R];\\ \text{Fl}_{t+R}^X(\gamma(-R)),&\text{if}\ t\in[-R-\epsilon,-R+\epsilon].\end{cases}$$
Since this procedure can be iterated indefinitely, one concludes that each maximal integral curve of $X$ is defined on the entire $\mathbb{R}$, i.e. the vector field $X$ is complete.
I hope this helps.
