First, note that the verb to expand has a specific meaning in this context: when you expand a node/state $s$, you try each action $a_1, \dots, a_n$ available from $s$, and each of these actions $a_i$ leads to another state.
Second, if the graph is a DAG, then the edges have a direction and there are no cycles; so, if you follow the edges, you will never get back to a node that you have already been at. However, this is not sufficient to understand that statement and why or when it is true.
Third, that statement is true only if you use the tree search version of the depth-first search, i.e. you don't keep track of the nodes that you have already explored. In other words, you don't use graph search, which is a tree search but that keeps track of the already expanded nodes in a set called the explored set (or closed list), so that to avoid expanding them again. So, here, if you explored a node/state $s$, you added it to the explored set, so you won't expand it again.
So, why is it possible to expand a node more than once in the tree search version of DFS? Well, you don't keep track of the nodes that you already expanded, so if you encounter it again from another path, you may expand it again. Given that it's a DAG (and assuming it's finite), then you will eventually try all possible paths, so you will find the goal.
Now, here's an example of when you expand a node more than once in the tree search version of DFS.
Let's say we have this search space, where you start at $A$ and the goal is $E$.
A -> B -> C -> D
A -> C -> D
A -> E
If the nodes are at the same depth, let's assume that you expand them alphabetically; so, in the search space above, you expand first $B$ and then $C$, because $B$ comes before $C$ (in the English alphabet).
Now, note that we can go directly to $C$ from $A$ or we can go to $C$ by first going to $B$. $C$ has a child. When you expand $C$, basically, you try the possible actions from $C$. In this case, there's only one action, i.e. going to $D$.
So, without keeping track that you expanded $C$, when you got there from $B$, after you backtracked from $D$ the first time, you will try again to expand $C$. In the graph search, this would not be possible because, the first time we expand $C$, we add it to the explored set, so we don't expand it again.