NP-complete reduction for a k-dumbbell graph
Hint: Following your idea, you want to reduce the clique problem to the dumbbell problem, to show that the latter is NP-hard. Given an instance $(G,k)$ of the clique problem, it is natural to consider two copies $G',G''$ of your original graph, connected in some way, so that if the original graphs contains a $k$-clique, then the new one contains a $k$-dumbbell.

If the new graph contains a $k$-dumbbell restricted to one of the copies, then the original graph certainly contains a $k$-clique. You want to arrange matters so that if the $k$-dumbbell straddles both copies, then it is composed of a $k$-clique in each one.

This is just a hint, so I'm not giving a full solution. Indeed, there might be several ways of using this idea to get an NP-hardness reduction. Some of them require you to slightly change the value of $k$ from the clique instance to the dumbbell instance.