I am working on a combinatorial proof given that $n-k$ is divisible by $g$ (that is $mod(\frac{n-k}{g})=0$). I am missing a piece for this step:
$\frac{k-1}{n-1}{n+\frac{n-k}{g}-2 \choose n-2}+\frac{k+g}{n}{n+\frac{n-k}{g}-2 \choose n-1}=\frac{k}{n}{n+\frac{n-k}{g}-1 \choose n-1}$
For convenience let $\ell=\frac{n-k}g$; the desired equality is then
$$\frac{k-1}{n-1}\binom{n+\ell-2}{n-2}+\frac{k+g}n\binom{n+\ell-2}{n-1}=\frac{k}n\binom{n+\ell-1}{n-1}\;.\tag{1}$$
We can use the identity $$\binom{n+\ell-1}{n-1}=\binom{n+\ell-2}{n-2}+\binom{n+\ell-2}{n-1}$$ to expand the righthand side of $(1)$ and rearrange it to get
$$\frac{k+g}n\binom{n+\ell-2}{n-1}-\frac{k}n\binom{n+\ell-2}{n-1}=\frac{k}n\binom{n+\ell-2}{n-2}-\frac{k-1}{n-1}\binom{n+\ell-2}{n-2}\;,$$
which immediately simplifies to
$$\frac{g}n\binom{n+\ell-2}{n-1}=\frac{n-k}{n(n-1)}\binom{n+\ell-2}{n-2}\;.$$
Now $n-k=g\ell$, so we can multiply both sides by $\frac{n}g$ to get
$$\binom{n+\ell-2}{n-1}=\frac{\ell}{n-1}\binom{n+\ell-2}{n-2}\;.$$
And
$$\frac{\ell}{n-1}\binom{n+\ell-2}{n-2}=\frac{(n+\ell-2)!}{(\ell-1)!(n-1)!}=\binom{n+\ell-2}{n-1}\;,\tag{2}$$
so all that you have to do to prove $(1)$ is start with $(2)$ and reverse the steps that I used to get from $(1)$ to $(2)$; they are all reversible.