Series current calculations
Question 1: How come I get 2 different answers when applying the 2 different ways to calculate I? Where am I going wrong here?


When using I = P_Z1/U_Z1, you assume to know the power dissipated by the resistor. But you don't know this yet. Like muyustan comments, 2W is the rating of the resistor, not the power dissipated at the moment.  

The power dissipated is a dependent variable. The known independent variables in your circuit are the resistance and the applied voltage.


  Question 2: In circuit 1 (and 2) I connected a 12V heating element to a 24V power supply and calculated I throug 2W/12v = 0.17A. I get the feeling that this is wrong because the source voltage is not taken into acocunt here, where am I going wrong here?


Although you take the source voltage into account (2W/12v = 0.17A), you cannot assume the power dissipation. See previous comment.


  Question 3: what does it practically mean when connecting a 12V/2W element to a 24V source?


First: It means you use a component beyond its rated values. You should not connect a 12V rated component to a 24V power source.
Next, if the component survives being subjected to a voltage higher than the rated voltage, it doesn't mean it will behave the same as when being used within the rated operating conditions.


  Question 4: In circuit 3 we have I = 0.14A. this means for Z1 (and Z2) that P has changed? P_z = 12V*0.14A= 1.68W, but when we started it was 2W. Where Am I going wrong here?


As written above, the power dissipation can change.
When "12V/2W" is the only information you have for that resistor, you can assume  the resistors has a power dissipation of 2W at 12V. This means its resistance is \$R=U^2/P = (12V)^2/2W = 72 \Omega \$. Likewise the 12V/1W resistor has a resistance of \$ 144 \Omega \$.  

Assuming the resistance does not change with applied voltage, even when it is beyond its rating, you can find the current through the resistors is
$$ I = 24V / (Z_1+Z_2) = 24V / 216 \Omega = 0.11 A$$

The voltage across Z_1 will be \$ 0.11A*72 \Omega = 8V\$
The voltage across Z_2 will be \$ 0.11A*144 \Omega = 16V\$ which is beyond its voltage rating.  

The power dissipated by Z_1 will be \$ (0.11A)^2*72 \Omega = 0.88W\$
The voltage across Z_2 will be \$ (0.11A)^2*144 \Omega = 1.76 W\$ which is beyond its power rating