Deciding between two algorithms with similar runtime in two parameters
A better comparison is $n$ versus $m\log m$. When $n = \Theta(m\log m)$, both algorithms have running times $\Theta(m\log m)$. When $n \gg m\log m$ (or rather $n = \omega(m\log m)$, the second algorithm has running time $\Theta(n)$ and the first one has running time $\Theta(m\log n) = o(n)$ since $m = o(n/\log n)$. When $n \ll m\log m$ (or rather $n = o(m\log m)$) the second algorithm has running time $\Theta(m\log m)$ and the first one has running time $\Theta(m\log n) = o(m\log m)$.