An example of handle decomposition on modified $S^5$ 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, below, seems quite complicated (then I would like to generalize the construction to $S^5$ with more copies of $S^1\times D^4$ removed, under certain combinatorial conditions).
Is the following attempt a valid handle decomposition of $M$ ?
Attempt to construct M:
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 believe 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, take one $D^4=D^1\times D^3$ of each handle and connect the $D_1$ of each two handles on one side to create a $S^1\times D^3$. Do the same on the other ends of the two 1-handles as well. The $D^1$'s of the co-attaching regions get connected as well to give a $S^1$. The result has boundary $(\sqcup_{2 copies} S^1\times D^3)\cup S^1 \times S^3$. Take the result and an another copy of itself, and connect the four disjoint $S^1\times D^3$'s two by two, giving an object with two disjoint $S^1\times S^3$ as boundaries; these are the boundaries of the target manifold. Note that at this point we did not do any gluings of handles but only combination of disks along their 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 $(D^5\times D^0,S^4\times D^0)$.
 A: I didn't really follow the details of your description, but here is a handle decomposition along the lines you suggest. Start with $((S^1 \times S^3)_a \coprod (S^1 \times S^3)_b) \times I$ where $I= [0,1]$. Add a 1-handle along $((S^1 \times S^3)_a \coprod (S^1 \times S^3)_b) \times \{1\}$. Note that by choosing the attaching region of the $1$-handle, you can make it miss a copy of $S^1 \times B^3$ in each of $(S^1 \times S^3)_a \times \{1\}$ and $(S^1 \times S^3)_b\times \{1\}$
Now add two 2-handles, one along each of the aforementioned copies of $S^1 \times B^3$. If you glue copies of $S^1 \times B^4$ along $(S^1 \times S^3)_a \times \{0\}$ and $(S^1 \times S^3)_b\times \{0\}$ then all of the handles cancel and you've got $B^5$ with boundary $S^4$.
It follows that if you add a $5$-handle, you get $S^5$ minus a particular pair of copies of $S^1 \times B^4$. But since any pair of circles with a given in $S^5$ is isotopic to any other pair, this is a description of any pair. The same description would work for any number of circles.
