Why does the collector current direction remain the same in saturation and active region?
The Ebers-Moll model actually considers this issue.

Having noted that it is not really possible to model a transistor as two diodes, it is possible to model it as two functions of the same transistor.

If you wish for a full canonical answer I can provide it, but I will try to stay intuitive at this point.

Start in the normal operating mode, where \$V{bc}\$ is \$\le \ 0\$ ( reverse biased) and \$V_{be}\$ is present and above the threshold; therefore the current gain is in the active region.

Now reverse the situation such that \$V_{bc}\$ is present (collector base forward biased) and \$V_{be}\$ is 0. This reverses the transistor and swaps emitter and collector, but due to the doping levels of a standard transistor, the current gain is much lower in this mode. (The gain is proportional to doping levels and the emitter is more heavily doped than the collector)

When superimposing the gains, the normal forward gain is still larger than the reverse current gain and therefore the overall current gain is still in the sign of the normal forward mode, but at a much lower value (which is why \$ \beta \$ is very low at low \$V_{ce}\$ and therefore why \$I_c\$ is very low at low \$V_{ce}\$; this implies that \$V_b \ is \gt V_c\$ for a NPN device).

The overall large signal current gain (and therefore the effective direction of current) is strictly given by:



The first term describes the first situation (normal forward bias outside of saturation) and the second term the reverse situation (collector > base for NPN); \$ \beta_R\$ is the reverse current gain.

There is an excellent thorough analysis available.