Let $\gamma(t)$ be the maximal integral curve with initial condition $\gamma(0) = m$, and let's suppose $b(m) < \infty$.
If we denote $\gamma'(t)$ the maximal integral curve with initial condition $\gamma'(0) = \gamma(b(m) -\epsilon/2)$ then
$$
\gamma''(t) =
\begin{cases}
\gamma(t) & \text{if } t\in (a(m), b(m) - \epsilon/2)\\
\gamma'(t - b(m) + \epsilon/2)& \text{if } t\in [b(m) - \epsilon/2, b(m) + \epsilon/2)
\end{cases}
$$
prolongs $\gamma(t)$. That contradicts the assertion that $\gamma(t)$ is maximal.
A similar reasoning can be used to show $a(m) = -\infty$.
Edit - Some clarifications
$\gamma''$ prolongs $\gamma$ means that $\gamma''$ has an interval of existence, $(a(m), b(m) + \epsilon/2)$, that is a proper superset of the interval of existence, $(a(m), b(m))$, of $\gamma$ and the two curves are equal on the common domain, $(a(m), b(m))$.
The "uniformity" of $\epsilon$ ensures that the integral curve with initial condition
$$
\gamma'(0) = \gamma(b(m) - \epsilon/2)
$$
exists on the interval $[0, \epsilon)$. Without "uniformity" it could happen that the maximal right interval of existence of $\gamma'$ is $[0, \epsilon/2)$. In such a case, a curve $\gamma''$, constructed as above, would coincide with $\gamma$: it would not prolong $\gamma$.
The initial condition of $\gamma'$ is chosen in such a way to allow us to smoothly join $\gamma'$ to $\gamma$ in order to form a new curve, $\gamma''$, that prolongs $\gamma$.
The contradiction arises because we constructed an integral curve that prolongs a maximal integral curve, which, as such, cannot be prolonged by definition.
Your argument is good. Here's how I'd pick appropriate $V_p$ and $t_p$:
Fix coordinates $(x,y^1,y^2,\ldots)$ on $U_p$ such that $\partial/\partial x = V$. Since $U_p$ is a neighbourhood of $p$, it contains some coordinate ball $B_r(p)$. Since flow solutions are unique, $\theta(t,x,y) = (x+t,y)$ is true whenever $(x,y)$ and $(x+t,y)$ are both in the coordinate chart domain; and by the triangle inequality we know $(x+t,y) \in B_r(p)$ whenever $|t|<r/2$ and $(x,y)\in B_{r/2}(p).$ Thus we can take $V_p = B_{r/2}(p)$ and $t_p = r/2$.
Best Answer
Let $M=\mathbb R$ and $X=(x^2+1)\frac{\partial }{\partial x}$. Then $M$ is complete, $X$ does not vanish, but every solution of $\varphi '=\varphi^2+1$ blows up in finite time, because it's of the form $\varphi(t)=\tan (t-C)$. Therefore $X$ is not complete.