Your current attempt (at the time of writing this answer at least) is on the right track, it just needs one more step.
To help understand one way of relating the problems to each other, consider the following alternative way of defining a vertex cover; a vertex cover is a set of vertices whose removal leaves a disconnected graph.
So Vertex Cover is the problem of selecting vertices to remove that gets rid of all edges, whereas your problem is selecting vertices to get rid of triangles.
Thus to convert Vertex Cover to Triangle Removal we need to turn edges in the original graph intro triangles in the new graph (or at least, one sensible way - there could be many reductions).
Hence your construction of adding a vertex for every edge in the graph and making a triangle is heading in the right direction, the only problem is that we could remove the new vertices, and that may mess up our answer (it certainly makes it harder to argue about what the vertex cover should be in the original graph). What we need is a way of preventing the added vertices being selected for removal. To do this, instead of adding a single new vertex, add $k+1$ new vertices!
More precisely (but not completely precisely), given $G$ we construct $G'$ as follows:
- For every vertex $v \in V(G)$, there is a vertex $v' \in V(G')$
- For every edge $uv \in E(G)$, there is an edge $u'v' \in E(G')$.
- For every edge $uv \in E(G)$, there are $k+1$ vertices $\{w_{1},\ldots,w_{k+1}\} \subseteq V(G')$ with the edges $u'w_{i}$ and $v'w_{i}$ in $E(G')$ for all $i$.
I won't go through the full proof that this works, but the general gist is that if we want to get rid of all the triangles attached to $u'v'$, it doesn't really help to take any of the $w_{i}$ vertices as there are too many of them, we eventually have to take at least one of $u'$ or $v'$, but once we take one of those, it deals with all the triangles, so any answer that takes a $w_{i}$ vertex must also take $u'$ or $v'$ (or both) and can be replaced by a smaller solution that takes only $u'$ or $v'$ (or both), so we only have to consider solutions of that form, but then we're basically done - as each edge has triangles, one of the two "original" vertices must be in the triangle removing solution for each edge, and hence is also a vertex cover.
Really you can make the same argument with your construction, but the additional vertices force things and make the argument clearer (you don't have to deal with arguing for alternative solutions and vertex swapping etc. etc.)
