however there is no collector base voltage that we must overcome since as soon as we apply 0.7 volts to the base the transistor already conducts between emitter and collector without offering resistance
That is correct.
For a full explanation, please consult a text on semiconductor theory. What I present here is a simplification which I hope is accurate enough.
Inside a diode there are three regions, an N-region, a P-region, and at the junction between them, a depletion layer. If the voltage between the N-region and the P-region is insufficient, very few electrons in the N-region cross the depletion layer, and very few holes in the P-region cross the depletion layer. When there is enough voltage between the N-region and P-region, electrons will cross from the N-region to the P-region, and holes will cross from the P-region to the N-region.
When electrons cross over from the N-region to the P-region, they do not immediately combine with holes, but continue travelling for some distance as "minority carriers". They are minority carriers because they are electrons in a P-regions. Similarly, when holes cross over from the P-region to the N-region, they do not immediately combine with electrons, but continue travelling for some distance as minority carriers.
If the N-region is more heavily doped than the P-region, then more electrons will cross over from the N-region into the P-region, than holes will cross over from the P-region to the N-region. Whether it is mostly electrons crossing the depletion region or mostly holes crossing the depletion region doesn't really matter in a diode. The diode is conducting current either way.
For the sake of simplicity, in what follows, when I refer to a "transistor", I will mean an NPN transistor. What happens in a PNP transistor is similar except that the roles of electrons and holes is swapped.
When the base-emitter junction of an (NPN) transistor is forward biased, electrons cross over from the N-region emitter to the P-region base. There, they are minority carriers. They don't recombine with holes in the base immediately. If the base is thin, and the average path of an electron before recombining is long enough, the electrons will reach the base-collector junction. Because they are minority carriers, they pass over the base-collector junction easily, and enter the collector. Once in the collector, they become majority carriers again, and travel freely to the metal collector terminal. That is why, when the the emitter-base junction is forward biased, current will flow from emitter, through base, into collector, even when the base-collector voltage differential is less than that of a typical diode junction or reverse biased.
The doping concentration of the collector is significantly lower than that of the base or emitter. So, for an NPN transistor, electrons easily flow in the direction just described. It is also possible for the roles of the emitter and collector to be reversed, but because the doping concentrations in the emitter and collector are significantly different, a transistor biased in "reverse active mode" doesn't have the same electrical characteristics as one in "forward active mode". Because the base is more heavily doped than the collector, it is much easier for holes to cross from the base into the collector, (when the base collector junction is forward biased) than for electrons to cross from the collector into the base. That is why the collector of a transistor does not make a good "emitter", hence why the reverse-active mode is not as widely used as the forward active mode.
The take away: When electrons cross from the emitter N-region emitter into the P-region-base, they do not recombine with holes immediately. They become minority carriers. As minority carriers, they may cross over the the base-collector junction even when or even though the collector may be at a lower voltage than the base.