Majority carriers (electrons for N type semiconductors and holes for P type semiconductors) do not significantly cross a PN junction until a characteristic forward voltage (dependent upon the material and temperature) is reached. For silicon diodes, this is typically around 0.6-0.7 volts. In contrast, minority carriers (electrons in a P type region, or holes in N type region) cross a PN junction with relative ease.
When majority carriers cross a PN junction, they become minority carriers, i.e. electrons in a P type region, or holes in an N type region. Likewise, when minority carriers cross a PN junction, they become majority carriers. So, if a majority carrier crosses a PN junction, and becomes a minority carrier, and if it arrives at another PN junction (which will, logically need to be oriented in the opposite way to the first), then that minority carrier will easily cross the second PN junction (and become a majority carrier again.) The fact that it will easily cross the second junction is the basic principle behind bipolar junction transistors.
At this point, I would like to make a tangent point that is not strictly necessary for an answer to your question, but is interesting nonetheless. I wrote above that if a majority carrier crosses a PN junction, and becomes a minority carrier, and if it then reaches a second (reversely oriented) PN junction, then it will cross that second junction easily. However, if a majority carrier crosses a PN junction and becomes a minority carrier, and if that minority carrier now finds itself in a vast sea of majority carriers, the chances are good that the minority carrier and a majority carrier will meet and recombine. Obviously, once a minority carrier recombines, it will not continue its journey to a second (reversely oriented) PN junction. Thus, BJT transistor designers typically 1) dope the emitter much more heavily than the base, so that the "sea" of majority carriers in the base is not all that concentrated with majority carriers, and 2) make the distance between the emitter-base junction and the base-collector junction fairly small, so that the "sea" of majority carriers is not that "vast". These two design factors increase the percentage of majority carriers from the emitter that cross over the emitter-base junction and become minority carriers in the base, and then cross over the base-collector junction and become majority carriers in the collector. 99% efficiency in this regard is not atypical.
In a normal diode, even when there is not a sufficiently forward voltage for a significant number of majority carriers to cross the PN junction to generate a concentration of minority carriers on the other side, there is nevertheless always a certain number of minority carriers that are formed "spontaneously" on each side of the junction due to thermal effects. As mentioned, these minority carriers can pass easily across the PN junction, even when the PN junction is only slightly forward biased, but importantly, even when the PN junction is reverse biased. The flow of minority carriers across the PN junction when the junction is reverse biased accounts for most of the leakage current of a diode. The leakage current is small because the concentration of minority carriers is small. However, in a transistor, when a great deal of minority carriers are injected into the base from the emitter because the emitter-base voltage is sufficiently forward biased, the current across the base-collector junction can be large.
There are a number of ways saturation of a BJT has been defined. One commonly used definition is that a BJT is in saturation if both the base-emitter and base-collector junctions are forward biased. For an NPN transistor, this will be true when 0 < Vce < Vbe.
Now, why does the base-collector junction conduct even when the voltage across it is less than the typical 0.6-0.7 volts required for a diode to conduct?
The reason is that, just like the case where the transistor is the forward active region and the base-collector is reverse biased, when the base-emitter junction is sufficiently forward biased to conduct significantly, the base gains lots of minority carriers. Whether or not the base-collector junction is forward biased, or reverse biased, these minority carriers (or at least a large fraction of them) diffuse to the base-collector junction and cross it. When the base-collector junction is forward biased, but not sufficiently to cause significant majority carrier current, this base-collector current corresponds to the very small current that exists through a normal diode that is forward biased, but not sufficiently for significant majority carrier current. However, because there is more minority carriers in the base of a BJT in which the emitter-base junction is forward biased, than in a simple diode, there is correspondingly more current into the collector. The injection of minority carriers into the base from the emitter, permits a large current to flow across the base-collector junction which is forward biased, but not sufficiently to cause majority carrier current.
[Note that if the base-collector junction is sufficiently forward biased, majority carriers in the base and collector will also cross over the base-collector junction and contribute to the collector current. However, if the base-emitter and base-collector junctions are both sufficiently forward biased for significant majority carrier current, then \$V_{CE}\$ is unlikely to be more than 0.2V and will probably be less.]
