4-manifold: $0$-handle $\cup$ $2$-handles along a framed link in $S^3$ (intersection form = linking matrix)

4-manifold: $0$-handle $\cup$ $2$-handles along a framed link in $S^3$
(intersection form = linking matrix)

Let $L$ be a framed link in $S^3$, consisting of framed knots
$L_1,\ldots,L_m$. Let $A=[a_{ij}]\in\mathbb{Z}^{m\times m}$ be its linking
matrix, with $a_{ii}=$ framing of $L_i$ and $a_{ij}=$ linking number of
$L_i$ and $L_j$.
Let $M$ be the $4$-manifold, obtained as a handle decomposition with one
$0$-handle $B^4$ and for every framed knot $L_i\subseteq S^3=\partial B^4$
one $2$-handle $B^2\times B^2$ glued along the frame $S^1\times B^2\approx
L_i$.
How can I prove that $A$ = intersection form of $M$ = presentation matrix
of $H_1(\partial M)$?
Here the intersection form of $M$ is defined as follows: $H_2M$ is
supposedly (don't know why) isomorphic to $\mathbb{Z}^m$ and generated by
immersions $f_i:kT^2\hookrightarrow M$ where $kT^2$ is the connected sum
of $k$-copies of the $2$-torus $S^1\times S^1$, and then the $ij$-th entry
of the intersection form is the intersection number of $f_i(kT^2)$ and
$f_j(lT^2)$, i.e. $\sum_{p\in f_i(kT^2)\cap f_j(lT^2)}\varepsilon(p)$
where $\varepsilon(p)=\pm1$, depending on what orientation the surfaces
$f_i(kT^2)$ and $f_j(lT^2)$ induce on $M$.
I can't even visualize the situation when $L$ is only one knot with its
framing $n$-times twisted, i.e. $M$ is $B^4$ with $B^2\times B^2$ attached
along $L\subseteq S^3$. What $3$-manifold is $\partial M$? $S^3$ with
$S^1\times B^2$ cut out and $B^2\times S^1$ pasted in with $n$-twisting?
Can't visualize that...