Let $M$ be a smooth manifold and $X \in \mathfrak{X}(M)$ be a vector field. If $X$ is complete, its flow defines a group action $\Bbb R \circlearrowright M$ via $t \cdot x \doteq \Phi_X(t,x)$. This gives us an action groupoid $\mathcal{G} \rightrightarrows M$, where $$\mathcal{G} = \{(t,x,y) \in \Bbb R \times M \times M \mid x = \Phi_X(t,y)\},$$with ${\sf s}(t,x,y) = y$, ${\sf t}(t,x,y) = x$, $M \hookrightarrow \Bbb R \times M \times M$ via $x \mapsto (0,x,x)$ and the operation is $(t,x,y)\cdot (s,y,z) = (t+s,x,z)$, everything standard here.
In Exercise 1.9 of Eckhard Meinrenken's notes, he wants to consider the case where $X$ is not complete, and the flow domain is just an open set $U \subseteq \Bbb R \times M$ (containing $\{0\} \times M$). The exercise asks us to show that $U$ can be turned into a Lie groupoid $U\rightrightarrows \Bbb R$ instead, apparently unrelated to the action groupoid described above.
It's completely unclear to me how to go about this. First because we're changing the manifold of units. Second because there doesn't seem to be a natural way to embed $\Bbb R \hookrightarrow U$. Third because things will go wrong with the operation above, since for $x \in M$ with $(t,x),(s,x) \in U$, one might not have $(t+s,x) \in U$ (duh).
Naively writing $$U = \bigsqcup_{x \in M} (a_x,b_x)$$and trying to define the source and target maps by relating $t$ in $(t,x)$ with $a_x$ and $b_x$ also seems bad, since we could have $a_x = -\infty$ and/or $b_x = +\infty$ even with $X$ incomplete.
Can you give me some ideas? I don't care for a full solution, I'm doing this for fun. If you tell me how to embed $\Bbb R \hookrightarrow U$ I'd say I already have a decent chance of figuring out the rest on my own. I don't know.
The source map is the projection $s: U\to M$, $s(t,q)= q$, the target map is the local flow $\Phi:U\to M$ of your vector field. The unit map sends $q\in M$ to $(0,q)$. The composition/multiplication in the groupoid is given by $$ (t,\Phi(s,q))\circ (s, q) = \Phi(t+s, q).$$
(Note that my convention is that the source of $(s,q)$ is $q$ and the target is $\Phi(s,q)$).