Just wondering if I could use two LM7805's in parallel to double the output current or would it be better to split the outputs to separate loads.
Related question: Is the heat dissipated dependent on current draw AND voltage reduced or just voltage reduced?
You can't parallel regulators like the 7805. And, the power dissipated (heat) will be the total current multiplied by the voltage DROPPED across the 7805. For example, if you use a 12 volt input to the LM7805 and you draw 1/2 amp out of it, the power dissipation will be (12-5) * 0.5 = 7 * 0.5 = 3.5 watts.
To get more output current, you can do one of 3 things:
(1) Use a different linear regulator (there are 3 amp and 5 amp and maybe higher versions of the classic 3 terminal regulator).
(2) Use a switching regulator.
(3) Use an external pass transistor to carry most of the load, like this:
(image courtesy of https://www.eleccircuit.com/boosting-regulator-current-for-ic-78xx-by-mj2955)
Note that you can get EXCELLENT remote sensing (if you need it) by tying the regulator output and the MJ2955 collector together at the LOAD end, as well as tying the 7805 ground lead to the ground rail at the LOAD end.
Hope this helps.