Borůvka cleanup in linear time?
This depends on what you mean by linear time. If you want $O(n)$, then you're out of luck, there's too many edges to consider in the worst case.

If you mean $O(n+m)$, then it is possible, and it is reasonable to consider this as linear time, as the input needs to represent the edges as well as the vertices, which takes at least $O(n+m)$ symbols.

After each round of selecting which new edges to add, we keep track of which pairs of components are being merged, then we look at these in order (any order is fine) and look at all edges from the new component to each other new component, and record the smallest we have seen. (Note that this is symmetric, so we when we find the smallest edge from new component $A$ to new component $B$, it is also the smallest from $B$ to $A$, so we only need to consider edges going to new components later in the ordering, but this doesn't affect the complexity). To achieve this we can simply look at all edges in order and update the appropriate information.