If your load dissipates 300 watts with 12 amperes through it, it'll require:
$$ E =\frac{P}{I} = \frac {300W}{12A} = 25 \text{ volts}$$
across it, and since your driving source can only supply 22 volts, your design will fail.
But it gets worse. ;) In the graphic below I've shown your setup using 11 AWG copper wire which has a resistance of 1.26 ohms per thousand feet, and 20 AWG, which has a resistance of 10.2 ohms per thousand feet.
Your load has a resistance of:
$$ R = \frac{E}{I} = \frac{25V}{12A} \approx 2.1\text{ ohms,}$$
and since resistances in series add, your source will be looking at (for AWG 11 wire):
$$ Rt = R1+R2+R3 = 1.26\Omega+2.1\Omega+1.26\Omega = 4.62\text { ohms}$$
Then, since current in a series circuit is everywhere the same, and since your source is limited to a 22 volt output, the current it can send into the string, thence into the load, is:
$$ I =\frac{E}{Rt} = \frac{22V}{4.62\Omega}\approx 4.8\text{ amperes}$$
With 4.8 amperes through your load, it'll drop:
$$ E = IR = 4.8A\times 2.1\Omega \approx 10.1\text{ volts}$$
and it'll dissipate:
$$P = IE = 4.8A\times10.1V \approx\text{ 48.5 watts}$$
Using 20 gauge wire makes the situation much worse, as shown below.

So what's the solution???
Using a buck converter connected to the load at the load, and driving the buck converter with a fairly high voltage from a source 1000 feet away might work, as suggested by others.
Just for grins, I've depicted, below, a buck converter running at 90% efficiency with a 300 watt output, a 100 volt DC input, and a variety of wire resistances connecting it to the source.
