Why the sign for \$4Ω\$ resistor is positive using KVL even though it dissipates heat?
I think you're a little confused, dissipation of power as heat isn't used anywhere in KVL. That comes into play when you're calculating things like efficiency and losses.

Firstly I think that question is laid out in a bit of a strange way. What they're trying to say is that the voltage drop across the \$8\Omega
\$ resistor is the same as the voltage drop across the \$4\Omega
\$ resistor. A more sensible way to write this would be \$I_{1}*8\Omega = I_{2}*4\Omega\$ or \$I_{1}*8\Omega - I_{2}*4\Omega = 0\$

The way it's written the way it is, is due to the position of the 'node' they've used (which kindly they haven't put on the circuit for us)



As for why the resistor doesn't have a negative sign, think about it in terms of a real world example. If you measure a voltage with a multimeter one way you get +12V, you reverse the leads you get -12V. Now if you do the same with a resistor you're going to get the same resistance either way.