trying to understand the mathematics behind point multiplication
The question is not performing any point multiplication, or any elliptic curve operation. It's exclusively performing arithmetic modulo the (prime) number $n$ of elements in the common elliptic curve group secp256k1, that is $n=$0xfffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364141. Every other integer in the question is an integer modulo $n$, not an elliptic curve point.
To get from an integer $k$ to an elliptic curve point (which the question does not attempt), we can use so-called point multiplication, multiplying that integer and some elliptic curve point. If $G$ is an elliptic curve point (including but not limited to the conventional generator point for secp256k1), then $k$ times $G$ is $[k]G=\underbrace{G+G+\ldots+G}_{k\text{ terms}}$ where $+$ stands for the operation in the elliptic curve group known as point addition#.
The group has order $n$, thus $[n]G$ is the group's identity element, also known as point at infinity, noted $\mathcal O$ or $\infty$. If follows that $[k]G=[k\bmod n]G$, and $[a]G+[b]G=[a+b\bmod n]G$, and $[a]([b]G)=[a\,b\bmod n]G$, which is why arithmetic modulo $n$ is of interest in the question.
Notice that for large $k$ we can do better than performing $k$ point additions to compute $[k]G$: for example we can compute $[9]G$ by computing $[2]G=G+G$, $[4]G=[2]G+[2]G$, $[8]G=[4]G+[4]G$, $[9]G=[8]G+G$. This allows to efficiently compute $[k]G$ for very large integers $k$, by reducing $k$ modulo $n$, then performing less than $2\log_2n$ point additions.
Going from point $P$ to point $Q=P+P$ is a special case of point addition, called point doubling. That's also multiplying $P$ by $2$ to get $Q=[2]P$. Going back from $Q$ to $P$ is point halving (a less common operation for secp256k1 at least). That's also multiplying by the modular inverse of $2$ modulo $n$, that is multiplying by $u=(n+1)/2=$0x7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a1.
If $Q=[42]G$, then halving $Q$ yields $P=[u]([42]G)=[u\times42\bmod n]G=[21]G$.
Similarly, if $Q=[43]G$, then halving $Q$ yields $P=[u]([43]G)=[u\times43\bmod n]G$. The question correctly computes $u\times43\bmod n=$0x7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20b6, which also is $(n+43)/2$. But the question is incorrect when stating that's the multiplicative inverse of 0x15.
how do I perform division
If that's in integers modulo $n$ (as in the question): in order to divide by $x$, multiply by the multiplicative inverse of $x$ modulo $n$. The question does just that for $x=2$, and successfully divides by $2$ first $42$ then $43$. Equivalently, to divide by $2$ an even integer modulo $n$, do just as for a normal integer. And to divide by $2$ an odd integer modulo odd $n$, add $n$ in the integers (thus yielding an even integer) then divide by $2$.
To divide an elliptic curve point $Q$ by integer $x$, compute $x'=x^{-1}\bmod n$, then compute $[x']Q$. We have an open question about how to perform that faster for $x=2$ and curves like secp256k1. Alternatively, if it's known $k$ and $G$ such that $Q=[k]G$, we can divide $k$ by $x$ modulo $n$ as in the above paragraph yielding $k'$, and the desired result is $[k']G$.
# The methods for point addition (which the question does not attempt) depend on the elliptic curve, it's underlying field, and the coordinate system. For secp256k1, it's used a prime field $\mathbb F_p$, thus arithmetic modulo the prime $p\ne n$ is involved. When using Cartesian coordinates, these standard formulas can be used.
