What blows the fuse is really the power wasted at it. \$p(t)=u(t)\cdot i(t)\$.
But the voltage here is not the supply voltage of the circuit, but the voltage drop across the fuse, which is determined by the current passed through it. \$p(t)=R\cdot i^2(t)\$. R is constant, so it really depends only on the current.
In typical setting, the resistance of the fuse will be negligible compared to resistance of the load, so the current running through it will be determined solely by the load resistance (or impedance in general).
In case of short circuit, there will be no load impedance and current will be limited only by the fuse's filament. In such case, supply with higher voltage will make it blow faster. (\$p(t)=\frac{u^2(t)}{R}\$)
Q1.1: The smaller the resistance, the higher the current passing through it. The resistor heated up so much, because the current was around I=6/10 = 0.6A P = VI = 6*0.6 = 3.6W which is probably much higher than the (assumed 1/4 or 1/2W) rating of the resistor. This resulted in the overheating and burning. (The calculation is estimate. I'm not sure what is the voltage drop on the peltier if there is any. The calculation assumes it is a short circuit, but probably it has some resistance.)
Q1.2: It doesn't make any difference. In a circuit where all components are in series, the passing current in all components will be the same. Basically all current goes into the circuit will come out at the end. The current will not be consumed. (Kirchhoff's current law).
Q2: Assuming that the peltier has its internal resistance, it should be generally no problem to connect it directly. However, if too much current flows through it, the peltier's hot side will heat up quickly faster than the heat sink can cool it and will heat up eventually the cool side as well. Should that happen, you will either need to control the current, reducing it or improve the cooling method e.g. increase the heat sink, put a fan on it or use water cooling.
To control the current you may want to either introduce a big resistor, but given the relatively high current, it should be a big power resistor which are not too cheap, or you should get a power supply which allows you to control the output current.
Best Answer
Energy (measured in kWh) is simply the sum of power over time (measured in Watts, or Volts * Amps).
So, if you use 600 kWh in one month (I'm assuming that's what you meant), you can simply divide energy by time to get average power. Google is actually great for these type of conversions (you get 821 Watts): https://www.google.com/search?q=600%20kwh%20%2F%201%20month
Now, since Watts are Amps * Volts, you can just divide your average power by your voltage (120 Volts if you're in the US) to get your average current. Going to google again, it looks like 6.8 Amps: https://www.google.com/search?q=600%20kwh%20%2F%201%20month+%2F+120+volts
Of course, this ignores the fact that some of your power is probably used at 220 volts, such as washing machines and some furnaces and water heaters, but hopefully it is helpful anyway.