Ratios of produced gases in water electrolysis
We know from stoichiometry that you did produce twice as many moles of $\ce{H}$ at the negative electrode as $\ce{O}$ at the positive electrode. However, we don't know a few important things:

What were the actual volumes of the gases that were finally collected? Volume is notoriously hard to estimate by eye, unless you are comparing the levels in two identical containers. So, we'll take your word for it that it was &quot;more&quot;, but we won't place any bets on the factor of 10.
What happened to the oxygen and hydrogen after the water split? Did it react?
What else was in the water that could have reacted? We know there is $\ce{NaCl}$, but maybe there are other dissolved compounds as well?

Question 1 can't really be answered here - the best we can do for now is assume that the volume of hydrogen was 2-10 times larger than that of oxygen. To find out for sure, you would need to conduct another experiment using graduated collection vessels.
Question 2 we can partially answer. If the electrode was corroded, then something reacted - either chlorine, oxygen, or some other more complicated set of reactions. We can get a rough estimate of the volume of oxygen this might consume by assuming it was oxidized by oxygen. If that 0.1 g of the electrode was oxidized, and that it was oxidized into $\ce{Al(OH)3}$, then at 1 atm and 298 K it would consume roughly 137 mL of $\ce{O2}$:
$\ce{Al^3+ + 3OH^- -&gt; Al(OH)3}$
$0.1 \space \rm{g \space Al} * \frac{1\rm{\space mol \space Al}}{27\rm{\space g \space Al}} * \frac{3 \space mol \space O}{1 \space mole \space Al} * \frac{1 \space mol \space \ce{O2}}{2 \space mol \space O} = 0.0056 \space mol \space \ce{O2}$
$PV = nRT$ (oxygen is close enough to ideal under normal lab conditions)
$V = \frac{nRT}{P} = \frac{(0.0056 \space \rm{mol} \space \ce{O2})(0.08206 \space L \space atm \space mol^{-1}K^{-1})(298 \space K)}{1 \space \rm{atm}} = 137 \space \rm{mL}$
$\ce{Al(OH)3}$ is a likely candidate for a reaction product because of the white flakes you observed - aluminum hydroxide is insoluble in water. Aluminum oxide ($\ce{Al2O3}$) is also a potential candidate, and would consume 68 mL of $\ce{O2}$ under the same conditions.
So, assuming 0.1 g is a reasonable estimate for the mass of corroded aluminum, it seems very possible that what happened is the oxygen reacted with the aluminum electrode, reducing your yield of oxygen gas.
How can you get around this? Well, my five minutes of google research did not turn up any good candidates for inert electrodes. Usually graphite electrodes are recommended for other electrolysis setups, but they will react with oxygen as well. I have seen stainless steel recommended, but I would think that gold/platinum/silver are probably your best bets.
Question 3 is also hard to answer. We know that the $\ce{NaCl}$ will react and produce hydroxide ion (further supporting our hypothesis for aluminum hydroxide formation) at low concentrations, and chlorine gas at higher concentrations. If it were producing chlorine gas, you would know - it would be a pale yellow/green and it would make you sick as soon as you smelled it. So it is unlikely that your concentration of table salt was high enough for that to be the competing reaction. However, it would contribute to the aluminum oxidation. The easiest way to rule this out is to not use table salt. I have seen baking soda (sodium bicarbonate) recommended instead, and that would probably work. However, it might still produce hydroxide in solution. What about other unknown substances? If you used tap water, it is likely that there is additional chloride, possibly fluoride, and probably various magnesium and calcium salts. All of these would react during the hydrolysis, and would affect yields. The way to eliminate them is to buy or make distilled or deionized water.
Summary:

Why did I get less gas than expected at the electrode connected to the +?
Was the oxygen consumed by oxidising the electrode? (It appeared quite damaged at the end.)

The answer is most likely yes, the oxygen reacted with the aluminum. I would bet that it formed aluminum hydroxide, which then precipitated into the white flakes you observed.

How can I get a ratio of 1/2?

You might not be able to get it to exactly 1:2, but if you use sodium bicarbonate or a weak acid (vinegar would probably work) instead of salt as an electrolyte, and an inert electrode - like gold, or platinum, you can probably get close. Keep in mind - hydrogen gas is usually the desired product in hydrolysis. Oxygen is cheap if you don't need it to be pure! And even pure oxygen is relatively cheap compared to other gases.