Sum rule for Big-O with equal complexity-functions?
It depends on if some variables on dependent on another, or if they are given as parameters in the problem. For example, in many graph-related problems, $n$ can be the number of vertices, and $m$ can be the number of edges. In this case, $m$ can be as large as $O(n^2)$. So in the general case, say if we have an algorithm whose first phase runs in $O(n)$ and second phase runs in $O(m)$, we just add the two terms together and do not attempt to simplify --- the final runtime is $O(n+m)$.
In terms of several variables given as parameters that aren't necessarily related: eaving multiple variables inside the final equation is standard, e.g. I can say that multiplying an $m\times n$ matrix and an $n\times p$ matrix takes $O(mnp)$-time using schoolbook multiplication, or solving the knapsack problem takes $O(nt)$-time where $n$ is the number of items and $t$ is the target sum.
In special cases like planar graphs and other sparse graphs, we know that $m=O(n)$, so we can safely substitute $n$ in place of $m$ in the final running time for simplification.
As a digression, this illustrates why on sparse graphs (where $m=O(n)$) one would prefer to use Dijktra's in a loop for all-pairs shortest path (in contrast to a transitive-closure based algorithm like Floyd-Warshall); since now Dijkstra runs in $O(m \log n) = O(n \log n)$, the overall complexity for using Dijkstra's in a loop becomes $O(n^2 \log n)$, which is better than Floyd-Warshall. By contrast, note that Dijkstra's runs in $m \log n$ for the general case, which means on dense graphs where $m=O(n^2)$, it becomes less efficient than Floyd-Warshall in terms of worst-case running time.
In other cases, there can be variables aren't even polynomially bounded by a given parameter --- take the knapsack example above, where $t$ can be exponential in $n$.