I don't have a reference, but I think it's not too hard to see that $M_1 \#_k M_2 \approx M_1 \# X \# M_2$, where $X = (S^1 \times S^{n-1})^{\# (k-1)}$. (Connected sum depends on choices of embedded disks, and so does $k$-connected sums; I'm assuming you want all $k$ pairs of disks to be isotopic.) So decomposing using $k$-connected sum is not so different from decomposing using usual connected sum, except for some bookkeeping about extra summands of $S^1 \times S^{n-1}$.
user171227
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