To count the ways of sorting $n$ objects into $m$ bins where there are $k_i$ objects in the $i$-th bin, we use the multinomial coefficient ${{k_1+k_2+...+k_m}\choose{k_1}}{{k_2+k_3+...+k_m}\choose{k_{2}}}... {{k_{m-1}+k_m}\choose{k_{m-1}}} =\frac{n!}{k_1!k_2!...k_m!}$, but sometimes we want to divide by $m!$ because we don't care about the order in which we fill the bins. There are $m!$ permutations of the bins, but if the bins are identical then the permutations are all equivalent, thus we calculate $\frac{n!}{m!k_1!k_2!...k_m!}$. I don’t understand why this same logic does not apply to the "balls in bins" frame of counting problems.
$n^k$ counts the ways to put $k$ distinct objects into $n$ distinct bins, so if we divide by the number of permutations of the bins, which is $n!$, then we should arrive at the case where the bins are identical. Yet the “distinct objects into identical bins” problem is solved by Stirling and Bell numbers, which look nothing like $\frac{n^k}{n!}$
Likewise, the problem of putting $k$ identical objects into $n$ distinct bins is solved with ${k+n-1}\choose{k}$ (stars and bars), so the “identical objects into identical bins” problem would be solved by $\frac{{k+n-1}\choose{k}}{n!}$. But that is clearly wrong since it potentially results in a noninteger.
Why does dividing by $n!$ not serve to control for distinctness of bins in these problems, and what then is the actual combinatorial interpretation (if any) of $\frac{n^k}{n!}$ and $\frac{{k+n-1}\choose{k}}{n!}$?
Your assertion in the first paragraph is not correct.
Let's try to find a meaning for $\frac{n^k}{k!}$
Consider the simple case of putting $4$ distinct objects into $3$ distinct bins
Using the format $\,$[ Lay down pattern ]$\times$[permute pattern], we get
$4-0-0: \binom 4{4,0,0}\frac{3!}{2!} =3$
$3-1-0: \binom 4{3,1,0} 3! = 24$
$2-2-0: \binom 4{2,2,0}\frac{3!}{2!} = 18$
$2-1-1: \binom 4{2,1,1}\frac{3!}{2!} = 18 \to \boxed{63}$
What, then do you think $\Large\frac{63}{3!}$ indicates ?
I can't attribute any useful meaning to $\Large\frac{n^k}{k!}$
And the same holds good for the stars and bars version of $\binom{4+3-1}{3-1} = 15$ placements
ADDED
The punchline is that the multinomial coefficient puts distinct (=labeled) balls into distinct (=labeled) bins with matching labels, [$k_i$ into box $i$, not into any box], that is why when we want to count all ways of putting distinct objects into distinct boxes,[=$n^k$] we need to multiply two multinomial coefficients, one for the numner of possible patterns, and one for permuting each pattern, ergo $\frac{n^k}{k!}$ doesn't have a useful meaning