Adding Big-O and little-o notation to get a little-o
No. You are misusing Landau notation.


$o(f) + O(g) \not\subseteq o(f + g)$ even if $f \in O(g)$. The best you can get is $O(f + g)$. If $g \in o(f)$, you get $o(f)$.
$\log k$ and $\log n$ are incomparable as long as you can not put $k$ and $n$ in some kind of relation.


Assuming that $k \in \omega(1) \cap O(n)$, you can show that both $o(n \log k) \subseteq o(n \log n)$ and $O(n \log (n/k)) \subseteq o(n \log n)$. Then, we would indeed deduce that the proposed algorithm would take $o(n \log n)$ comparisons, contradicting the known lower bound.