I don't know if this answer will be satisfying to you, but there is some general topology context that might change how you look at this question.
Firstly, if you want to handle pairs of anything-with-topologies, you can use the product topology. This applies equally well to anything you have topologies on, and often gives nice answers. For example, if $\mathbb R$ has the standard topology and the circle $S^1$ has its standard topology (induced from its embedding in the plane, say), then:
- The product topology on $\mathbb R\times\mathbb R$ is indeed the standard topology on $\mathbb R^2$
- The product topology on $S^1\times\mathbb R$ corresponds exactly to the standard topology on an infinite cylinder living in $\mathbb R^3$.
- The product topology on $S^1\times S^1$ corresponds exactly to the standard topology on a torus living in $\mathbb R^3$.
So if we had a reasonable topology on a collection of functions, then we could get a reasonable topology on pairs of functions via this same product-topology construction.
Now the question is: what is a reasonable topology on a bunch of functions $f:X\to Y$? Well, suppose the functions are (some of) the continuous maps between two (possibly identical) topological spaces (maybe nice ones like metric spaces or something).
One candidate would be to treat the space of these functions just like (infinite if $X$ is infinite) products of the space $Y$, but this is ignoring the topology on $X$. Another common one which takes that into account is called the compact-open topology. For a number of reasons (one is mentioned at this MSE answer), the compact-open topology is usually nicer/more appropriate.
Since you used the tag functional analysis, if we're looking at the continuous (aka bounded) linear maps on a Hilbert space $H$, then there are lots of topologies in use. And this MO answer mentions that on norm-bounded subsets, the "strong operator topology" agrees with the compact-open topology I mentioned above.