Paredes and Navarro state that
$$m + k \log m = O(m + k \log k)$$
This gives an immediate "tighter looking" bound for incremental sorting. That is, if a partial or incremental sorting algorithm is $O(m + k \log m)$, then it is automatically $O(m + k \log k)$, where the $k$ smallest elements are sorted from a set of size $m$. Unfortunately, their explanation is rather difficult for me to understand. Why does it hold?
Specifically, they state
Note that $m + k \log m = O(m + k \log k)$, as they can differ only if $k = o(mα)$ for any $α > 0$, and then $m$ dominates $k \log m$.
This seems to suggest they're talking about $k$ as a function of $m$ along some path, but it's very hard to see how $k = o(mα)$ plays into things, or even where to place the quantifiers in their statement.
There are various ways to define big-$O$ notation for multi-variable functions, which would seem to make the question difficult to approach. Fortunately, it doesn't actually matter exactly which definition you pick, as long as you make the entirely reasonable assumption that $m > 0$ and $k \ge 1$. That is, in the incremental sorting context, you assume that you need to obtain at least the first element from a set with at least one element.
Theorem
If $m$ and $k$ are real numbers, $m > 0$, and $k \ge 1$, then $$m + k \log m < 2(m + k \log k).$$
Proof
Suppose for the sake of contradiction that
$$m + k \log m \ge 2(m + k \log k).$$
Rearranging terms, $$k \log m - 2k \log k \ge m.$$
By the product property for logarithms, $$k \log m - k (\log k^2) \ge m.$$
By the sum property for logarithms, $$k (\log (m / k^2)) \ge m.$$
Since $k$ is positive, we can divide it out to get $$\log (m / k^2) \ge m/k.$$
Since $k \ge 1$, $k^2 \ge k$, so (since $m > 0$) $m / k \ge m / k^2$. Thus $$\log (m / k^2) \ge m / k^2.$$
The logarithm of a number is always less than that number, so we have reached a contradiction.
I suspect that the $2$ in the above theorem could be reduced somewhat, but that doesn't matter for the order of growth.