In an NPN in active mode Vcb must be so large (with collector positive) that the maximum number of electrons from the emitter, coming through the base, are sucked into the collector. Increasing Vcb hardly changes the current as the electron flow is at maximum already. The electron flow is determined by the BE junction!
In an NPN in saturation mode Vcb is smaller, so small that the flow of electrons is influenced by Vcb. This is the red part of the graph in Andy's answer, a small change in Vce (which is just Vcb + Vbe) will cause a large change in Ic. In saturation many electrons make it to the base instead of being pulled into the collector. That makes the base current larger and the collector current smaller. When the base current is kept constant that would result in a smaller collector current.
Note that there is no "opposing" current, the BC junction does not need to be fully forward biased to already influence Ic. It is more the fact that the BC junction isn't so far into reverse mode for it to be able to "pull out" all the electrons it possibly can out of the base region. So less electrons reach the collector resulting in a smaller Ic.