Finding a tree that approximates the distances and total weights
Your problem is solved in the following paper by Khuller, Raghavachari, and Young. They show that you can construct a tree in which distances from the root are stretched by at most $\alpha$ and the total weight of the tree is at most $1 + 2/(\alpha - 1)$ times the weight of the MST. So, with $\alpha=3$, you can get $2$ times the weight of the MST. The algorithm does a depth-first traversal of the MST, and  adds paths from the shortest path tree when necessary, roughly speaking maintaining a shortest path structure in the current graph, which consists of the MST edges and the edges added from the shortest path three. Check the paper for details.

As I mentioned in the comment, there are graphs in which the weight of the shortest path tree rooted at some vertex is greater than the MST weight by $\Omega(n)$. One example is a path of unit weight edges, and edges from the first vertex in the path to the $i$-th vertex of weight just slightly less than $i-1$.