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 $S^1 \times S^{n-1}$.

I don't have a reference, but I think it's not too hard to see that $M_1 \#_k M_2 \approx M_1 \# X_k \# M_2$, where $X_k = (S^1 \times S^{n-1})^{\# (k-1)}$. (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 $S^1 \times S^{n-1}$.

The idea of this formula $M_1 \#_k M_2 \approx M_1 \# X_k \# M_2$ is that gluing the first neck gives $M_1 \# M_2 \approx M_1 \# S^n \# M_2$; the remaining $k-1$ necks may then be attached to the middle $S^n$ instead of between $M_1$ and $M_2$, each of which creates an $(S^1 \times S^{n-1})$-summand.