First, D2 and R2 are completely unnecessary, and are not mentioned in your link, so why you have them is a puzzle. However, since they have almost no effect on your circuit, they aren't actually hurting anything.
Second, your power supply voltage is too low. I recommend a supply at least 10 volts over your battery voltage, and more is better. The reason will become apparent.
Now, as to your circuit performance. I suggest that you get a second meter, and monitor the battery voltage as you charge. You will see that, during charge, the battery voltage will slowly increase. Since the charge current depends on the difference in voltage between the supply and the battery, charge current will drop as the battery charges. This is made worse by the inclusion of the LED, which will typically drop about 1.5 volts at your current. So the voltage across the charge resistor (R1) will start out at about 1.5 volts for a battery voltage of 9 volts. An increase of battery voltage of only 0.1 volt will cause the resistor voltage to drop to about 1.4 volts, which will cause a change in current from about 15 mA (for a 1k resistor) to 14 mA, and the trend will continue as long as the battery voltage rises with SOC.
So the effect is perfectly normal, given the simple circuit you are using.
Unless you can come up with a good reason for them, get rid of D2 and R2 and recalculate R1. Increasing the supply voltage will decrease the change in charge current. For instance, if you increase the supply voltage to a nominal 20.5 volts, a 9 volt battery and 1.5 volts across the LED will give you 10 volts across R1. If the battery voltage rises 0.1 volts, the resistor voltage will again drop by 0.1 volts, but this is only 1% of the original, and charge current will only drop by 1%.
But also keep in mind that your charging technique is not precise. Using exactly C/10 for a charge current, for exactly 10 hours, will not produce exactly 100% charge, and using a little more or a little less current will not have much effect on the final result.