Why do the two relations of Power in a resistive load $P= I^2 R$ and $P = V^2/R$ (Assume Ideal resistances ) seem contradicting in this situation?

I have seen similar questions in the forum but they fail to be very specific . I would ask the question from the viewpoint of Problem Solving. Let me put this question from a textbook :

A heater coil is cut into two equal parts and only one part is now used in the heater. The heat generated will now be?

Now from Ohm's Law, $$R \propto l$$, where $$l$$ is the length of conductor. So, since length is halved, $$R$$ is halved.

Since there is no mention in the question that the supply is Voltage Limited or Current limited, I assumed that the supply is a standard household supply which is voltage limited and I used the Relation $$P=\frac{V^2}{R}$$ and the fact that Heat generated is $$H= P\cdot t$$, where $$t$$ is time.

I got my answer correct which is,

The heat generated would be doubled

Now, the real question begins: What if we used the relation $$P=I^2 R$$, assuming supply to be current limited? Then the answer would be:

The heat generated would be halved.

Why such a contradiction arise? Is my assumption correct or is there some mistake that just somehow lead to a right answer ?

You just answered your question yourself: if the supply is voltage-limited, the first answer is correct. Under the assumption of constant voltage, what the expression $$P = I^2 R$$ really means is $$P = I(R)^2 R.$$ That is, current is a function of the resistance. Since $$I(R) = V/R$$ by Ohm's law, you get back $$P = V^2/R$$ by substitution. Of course, the conceptual opposite occurs if the supply is current-limited instead -- all your expressions in that case would reduce to $$P = I^2 R$$ with voltage being a function of the resistance.
It's not a contradiction! You're just considering two different cases, since it makes a huge different wether you have fixed current or fixed voltage. Now the thing is that if you have fixed voltage, you'd have to reinterpret the current intensity in the second answer as $$V=I\cdot R \Rightarrow I=\frac{V}{R}$$ and so, as you decreased $$R$$ by a factor two (by cutting the lenght of the coil), when calculating $$P=I^2\cdot R$$, you'd have a factor 4 multiplying coming from the current squared and a factor $$\frac{1}{2}$$ from the $$R$$ term, ending up again in a doubled power.