The collector current in your circuit decreases because, when the BJT is in saturation, by definition, the junction B-C is forward biased.
Therefore you'll have that the collector voltage will be just \$V_C=V_B-V_{BC}\$.
As a result, the current in the resistor Rc will be:
$$I_C=\frac {V_{CC}-V_C}{R_C} = \frac{V_{CC}-(V_B-V_{BC})}{R_C}$$
Therefore the higher \$V_B\$, the smaller \$I_C\$.
To answer the second question, simply use KCL at the BJT. You know that \$I_C+I_B-I_E=0\$ (assuming base and collector currents entering the BJT. Emitter current exits the BJT). Hence \$I_B=I_E-I_C\$. In fact in your plot, \$I_B\$ is the difference of the other two currents.
As a first order approximation, you can assume constant \$V_{BC}\$ and \$V_{BE}\$. This allows you to calculate \$I_C\$ and \$I_E\$ easily.