What's the approximation factor of this Max k-Cut approximation?
This is called the edge-contraction heuristic, in which an upper bound $(k-1)/(k+1)$ on the approximation ratio can be shown. See section 3 of the work by Kahruman et al. for reference.
Imagine whenever we combine two nodes u, v into a group, instead of forming a group we contract the edge (u, v) between these two nodes, and updates the weight of all the edges incident to the newly formed node. Repeat the procedure until there are only k nodes left. Lemma 3.1 in the above paper shows that the sum of weights of the first $i$ edges being contracted (denoted as $W_i$) satisfies the following inequality:
$$W_i \leq \frac{2iW}{(n-1)(n-i-1)} \text{,}$$
where $W$ is the sum of weights of all edges.
This can be derived from
$$W_{i+1} \leq W_i + \frac{W-W_i}{({n-i\atop 2})} \text{,}$$
since we always contract the lightest pair, the weight of the contracted edge cannot surpass the average weight.
One can see that the cut formed by this algorithm has weight $W_C = W-W_{n-k}$,
and by observing that $W \geq W^*$ where $W^*$ is the weight of the optimum cut,
we have our desired approximation ratio:
$$W_C \geq W-\frac{2(n-k)W}{(n-1)(k-1)} \geq \frac{k-1}{k+1}W^* \text{.}$$
For k=2, the algorithm gives a 1/3-approximation to the Max cut problem.