Transistor saturation current
Just to make notes about my assumptions before talking:


Those unlabeled curves are specified for \$I_\text{B}\$ set to specific values. This means the base current is fixed for each curve and doesn't change.
The value of \$V_\text{CE}\$ is the \$x\$-axis and \$I_\text{C}\$ is the \$y\$-axis.


Your question then is how to interpret what you see there.



How does \$V_\text{CE}\$ decline?

I suspect that you imagine there is a constant current supply used at the base and that there is also a voltage supply on the bench where someone just twiddles a knob on the voltage supply to change the voltage on \$V_\text{CE}\$, plotting the current that the voltage supply (or a meter) reports.

I suppose one could try it that way. But it doesn't help you understand what is going on nearly as well as thinking about this differently. So I'll put it to you in a different way and see if that helps, better.

Suppose that instead of varying a voltage supply tied directly across the emitter and collector, there still is that constant current supply at the base but now a fixed voltage (say \$10\:\text{V}\$) supply value is set up and there is a variable resistor used between this voltage supply and the collector. (If NPN, the emitter is grounded; or at the negative supply rail.)

The value for \$V_\text{CE}\$ is measured with a voltmeter attached appropriately, as the resistor value is varied. Assuming that \$\beta\$ is roughly fixed when in the active mode, the value of \$I_\text{C}\$ should not change as the resistor is varied. But \$V_\text{CE}\$ will vary because the resistor will drop a voltage that varies with its resistance. So this means we can still plot \$I_\text{C}\$ vs \$V_\text{CE}\$.

We expect to see:

$$V_\text{CE}=10\:\text{V} - R_\text{POT}\cdot I_\text{C}$$

But what happens when this varying resistance gets so large that the second term in the above equation, \$R_\text{POT}\cdot I_\text{C}\$, is close to \$10\:\text{V}\$? Certainly, it cannot be possible that \$V_\text{CE}\$ goes negative, right! There's no way for that to happen.

What starts to happen is that \$V_\text{CE}=10\:\text{V} - R_\text{POT}\cdot I_\text{C}\$ approaches the voltage at the base (which is probably somewhere between \$600\:\text{mV}\$ and \$900\:\text{mV}\$ for a small signal device.) As the voltage drop represented in the second term there forces \$V_\text{CE}\$ below that base voltage, the base-collector junction becomes forward-biased and starts to conduct significant current of its own, which can become quite large as the collector voltage approaches the emitter voltage very closely.

Keep in mind that we are using a current source at the base. So this newly developing base-collector forward-biased current doesn't add to the base current. It's important you hold that fact in your mind. There's a current source there, so that's just not possible to change in any way. Instead, this newly developing forward-biased base-collector current takes away from the recombination current that is required to allow collector current to flow.

The base current being supplied has now fractured into two meaningful parts: (1) \$I_{\text{F}_\text{BC}}\$, the forward-biased BC diode current; and, (2) \$I_{\text{F}_\text{RECOMB}}\$, the necessary recombination current to allow collector currents to flow.

Now \$I_\text{B}=I_{\text{F}_\text{RECOMB}}+I_{\text{F}_\text{BC}}\$, so obviously:

$$I_{\text{F}_\text{RECOMB}}=I_\text{B}-I_{\text{F}_\text{BC}}$$

Or, put another way, the newly developing forward-biased base-collector diode current takes away from the available recombination current that permits collector current to flow. So the collector current declines as a result.

And the larger this forward-biased base-collector diode current becomes (the larger the forward-biased base-collector voltage becomes) the deeper into saturation is the transistor as the remaining recombination current rapidly declines. "Shallow saturation" means that the forward-biased base-collector diode current is just a small portion of the total base current. "Deep saturation" means that the forward-biased base-collector diode current is a large portion of the total base current.

(This also happens if you apply that voltage supply directly across the collector and emitter, since this also means that there is a forward-biased base-collector diode current that is subtracting from the available base current.)