You can do this to drive your motor. The 1k resistor must be tuned based on your motor's expected current for a given voltage. Usually you have a spec like 500mA@12V or something of the like.
Basically, to get the 500mA spec, you want collector current of 500mA. The base current is always Ib=Ic/Hfe.
Tip122's Hfe is min 1000, so you get 500µA for Ib. You know on a BJT that Vbe is 1.3V for Tip122 at 500mA (see plot in datasheet). So if you have a GPIOs voltage of Vgpio, you have a voltage of Vgpio-2.5V across R1.
Using Ohm's law you can find out R1 => R1=(Vgpio-Vbe)/(Ic/Hfe).
For a 3.3V Gpio R1=(3.3V-1,3V)/(500µA) = 4K
For a 5V Gpio R1=(5V-1.3V)/(500µA) = 7.4K
Those resistor values are upper bound values to get your full 500mA. You should be using values around those.
simulate this circuit – Schematic created using CircuitLab
Now for your question about what is going on, I can't answer. It is not very clear about what you did and you look like you are mixing thing up a little.