I would like to give the following object, $M=S^5 \setminus \sqcup_{2 \text{ copies}} \text{int}(S^1\times D^4)$, a handle decomposition. It is then to be attached to another manifold. along the two copies of $S^1\times S^3$, so there is no need to start with a $0$-handle. My attempt seems quite complicated. Then I will generalize to $S^5$ with more copies of $S^1\times D^4$ removed, under certain combinatorial conditions. I lack an understanding of the rules for creating such object, so I am asking you if the following proposition is valid, or what you would do instead.

Start with two free floating $1$-handles $(D^1\times D^4, S^0\times D^4)$. They have co-attaching region $\text{co-att}=D^1\times S^3$. I would think that I am not allowed to decompose $D^4=D^1\times D^3$ and connect two $D^4$ of each handle in order to create a $S^1\times S^3$. In that case only connect the two handles on one side to create a $S^1\times D^3$ and on the other ends as well. Take the result, which has boundary $(\sqcup_{2 copies} S^1\times D^3)\cup S^1 \times S^3$ and another copy of itself, and glue the four disjoint $S^1\times D^3$'s two by two along $S^2$'s, giving an object with two disjoint $S^1\times S^3$ as boundaries. There are also two unwanted boundaries from the co- attaching regions $\text{co-att}$, which are $\#_{2 copies} S^1\times S^3$. Fill part of this two spaces by attaching two $2$-handles $(D^2\times D^3,S^1\times D^3)$. The result are two copies of $D^4$, one for each "holes". Now the fundamental group is trivial, and we can close the remaining unwanted boundary with a 5-handle $(D^5\times D^0,S^4\times D^0)$.