What is the best way to merge cycles to minimise total weight?
This is NOT an answer. This post is just to construct a concrete numeric example out of the left diagram in the picture in the question. 

(This answer, too long to fit as a comment, was requested by a user who wanted to see a concrete numeric example which his greedy algorithm may fail. It was written about the same time when OP added a numeric example that is exactly the same except a slightly different weight assignment.) 

Let $G$ be the complete undirected graph on 12 vertices $A_1, A_2, A_3, A_4, B_1, B_2, B_3, B_4, C_1, C_2, C_3, C_4$.

Let $S$ be the set of the following three simple circles.


the circle $c_A$: $A_1, A_2, A_3, A_4, A_1$.
the circle $c_B$: $B_1, B_2, B_3, B_4, B_1$.
the circle $c_C$: $C_1, C_2, C_3, C_4, C_1$.


All edges in the three circles weigh 1. The edges $A_3B_2, A_4B_1, B_3C_2, B_1C_4, C_3A_2$ and $C_4A_3$ weigh 2. 
  - All other edges weigh 16. 

We are allowed to merge two cycles by removing an edge from each cycle and connecting them up with two new edges, repeatedly until we reach a single cycle. The goal is to reach a cycle with the smallest total weight.

A description of an optimal solution for $S$.


Remove edge $A_2A_3$ and $B_2B_3$. Add edge $A_3B_2$ and $A_2B_3$.
Remove edge $A_2B_3$ and $C_2C_3$. Add edge $B_3C_2$ and $C_3A_2$.


We end up with the simple cycle $A_1, A_4, A_3, B_2, B_1, B_4, B_3, C_2, C_1, C_4, C_3, A_2, A_1$, which has weight $3* 2+ 9*1 = 15$.

A description of greedy algorithm on $S$ that considers all pairs of two existing cycles at a time.


Since the three cycles are symmetric, WLOG, let us merge cycle $C_a$ and cycle $C_b$. To avoid adding an edge whose weight is 16, we have only one choice. Remove edge $A_3A_4$ and $B_1B_2$. Add edge $A_3B_2$ and $A_2B_3$. The resulting circle, named $c_{AB}$, is $A_1, A_2, A_3, B_2, B_3, B_4, B_1, A_4, A_1$.
Now it is time to merge $c_C$ and $c_{AB}$. To avoid adding an edge whose weight is 16, we have only two choices. We can either remove $B_3B_4$ and $C_1C_2$ and add $B_3C_2$ and $B_4C_1$, resulting with cycle $A_1, A_2, A_3, B_2, B_3, C_2, C_3, C_4, C_1, B_4, B_1, A_4, A_1$, whose weight is $4 * 2 + 8 * 1 = 16$. Or we can remove $C_3C_4$ and $A_1A_2$ and add $C_3A_2$ and $C_4A_1$ cycle, resulting with cycle $C_1, C_2, C_3, A_2, A_3, B_2, B_3, B_4, B_1, A_4, A_1, C_4, C_1$, with the same weight 16.


Conclusion: A simple greedy algorithm does not guarantee the desired smallest weight.

I have also constructed a planar graph out of that left diagram. However, it is much more messier to describe the construction and to show the algorithms and their results.

Furthemore, it is not hard to construct an example where the greedy algorithm that considers all triples of existing cycles at a time will not reach the smallest weight. I believe, with a bit of more work, we can show a greedy algorithm that considers all combinations of a fix number of existing cycles does not guarantee the desired smallest weight.

It is quite possible this problem does not have a polynomial solution that always yields the smallest weight. In fact, it looks like this concern is implied by the OP since the question asks for "an efficient heuristic that produces a cycle of small weight".