We could model the twisted and untwisted sections like this:
simulate this circuit – Schematic created using CircuitLab
Each \$R\$ is some arbitrary length of wire, each length having the same resistance. The blue box represents a section of wire pair twisted together. The orange section is a length of single, untwisted wire.
Ignore for the moment the cross connections, such as the connection between A and B.
Current \$I\$ splits into two halves, shared equally between the two paths of the twisted section. Since \$P=I^2R\$, the power dissipated in each blue \$R\$ section is one quarter of the power dissipated in orange \$R\$. This implies that each path of the entire twisted section is receiving 25% of the heating power being received by the same length of untwisted section. The twisted pair together receive 50%. The twisted section is going to heat far less than untwisted section, with clear benefits for whatever mechanical connection is joining this wire to the power source at node C.
By twisting bare, un-insulated sections of wire together, obviously there will be points where the wires touch, represented by the cross connections such as the one between nodes A and B. Little or no current will pass via those points of contact, because the potential at each corresponding node on opposite sides (for example \$V_A\$ and \$V_B\$) will be very similar. Symmetry of the system ensures that potential changes at the same rate along both paths, and these cross connections have no consequence.