kJ per mole or kJ per 2 moles?
the whole reaction, i.e. for 2 moles of the product?


This is your mistake.

The balanced equation $\ce{2H2 + O2 -&gt; 2H2O}$ does not mean that 2 moles of the product are being formed.

The numbers, called stoichiometric coefficients, only tell you about the ratio in which the reactants react and form products.

That is to say, if I had a system in which I reacted 1 mol of $\ce{H2}$ with 0.5 mol of $\ce{O2}$ to form 1 mol of $\ce{H2O}$, the above equation would still be correct. This is because the ratio of $\ce{H2}:\ce{O2}:\ce{H2O}$ is still $2:1:2$. As long as the ratio is $2:1:2$, the equation will still be correct and applicable.



What $\Delta_\mathrm{r}H^\circ$ tells you is: how much the enthalpy change would be, if the number of moles reacting were the same as the stoichiometric coefficients. Therefore for the above reaction, saying that $\Delta_\mathrm{r}H^\circ = -572~\mathrm{kJ~mol^{-1}}$ means that

If 2 mol of hydrogen were to react with 1 mol of oxygen to form 2 mol of water, the enthalpy change would be -572 kJ.

That means that if you multiplied the equation by 2, $\Delta_\mathrm{r}H^\circ$ would also be multiplied by 2.

$$\ce{4H2 + 2O2 -&gt; 4H2O} \qquad \Delta_\mathrm{r}H^\circ = -1144~\mathrm{kJ~mol^{-1}}$$

This line tells you: if 4 mol of hydrogen were to react with 2 mol of oxygen to form 4 mol of water, the enthalpy change would be -1174 kJ. That's obviously true because you're reacting twice as much reactants to get twice as many products, so the enthalpy change must be twice as much.


  So what is the per mol for


As chemguide said, it is not per mol of anything. The per mol arises from the definition of the reaction enthalpy change, $\Delta_\mathrm{r}H^\circ$. I have another answer describing it in the context of Gibbs free energy here.

However, one way to look at it is to remember that the reaction does not stipulate the actual amounts of reactants and products. That is to say, the value $\Delta_\mathrm{r}H^\circ$ on its own cannot help you find an enthalpy change, unless you tell it exactly how much of the reactants and products you have. If you don't specify whether you start with 2 mol or 1 mol of water, you cannot find $\Delta H$ for the process.

That means you need to multiply $\Delta_\mathrm{r}H^\circ$ by an amount of substance (units mol) to get $\Delta H$ (units kJ). So, $\Delta_\mathrm{r}H^\circ$ must have units of kJ/mol.

What is the actual amount of substance that you are multiplying by then? It is simply the amount of substance of any species reacting, divided by its stoichiometric coefficient in the equation.

Therefore, if you have 6 moles of oxygen reacting, the "amount of substance" that you multiply by would be $6~\mathrm{mol}/1$. The enthalpy change would be 

$$\Delta H = (-572~\mathrm{kJ~mol^{-1}})\left(\frac{6~\mathrm{mol}}{1}\right) = -3432~\mathrm{kJ}$$

You have to divide by the stoichiometric coefficient. This is necessary for consistency. For example, if you were only given this $\Delta_\mathrm{r}H^\circ$:

$$\ce{4H2 + 2O2 -&gt; 4H2O} \qquad \Delta_\mathrm{r}H^\circ = -1144~\mathrm{kJ~mol^{-1}}$$

then the "amount of substance" you would multiply by is $6~\mathrm{mol}/2$. You would have

$$\Delta H = (-1144~\mathrm{kJ~mol^{-1}})\left(\frac{6~\mathrm{mol}}{2}\right) = -3432~\mathrm{kJ}$$

Note how the answer comes out to be the same.

But it doesn't have to be for oxygen either. That's what Chemguide means when it says it's not for any particular species in the reaction, because it's applicable to all of them. For example, if I told you that 44 mol of water was formed, then you would use

$$\Delta H = (-1144~\mathrm{kJ~mol^{-1}})\left(\frac{44~\mathrm{mol}}{4}\right) = -12584~\mathrm{kJ}$$