If I roll two fair dice, the probability that I would get at least one 6 would be....

That's right. The easier approach would be to calculate the chance of not rolling a $6$ - that's just $\frac56$ for the first die, and $\frac56$ for the second die, so by the product rule (as the events are independent), the probability is $\frac56 \cdot \frac56 = \frac{25}{36}$.

Then the probability of rolling a $6$ is $1$ minus the probability of not rolling a $6$, which we just calculated: so it is $1-\frac{25}{36}=\frac{11}{36}$.

If calculating the probability of an event not occurring to calculate the probability of it occurring feels weird to you, you might want to read up on complementary events. The idea here is that the product rule can sometimes make probabilities smaller when that wouldn't make any sense --- in our example, if you had just multiplied $\frac16 \cdot \frac16 = \frac1{36}$, then that would've been obviously wrong. This is because when we work with probability, we're dealing with quantities in the $[0,1]$ interval, so multiplication usually makes things smaller, rather than larger.

Yes indeed, you've got them all. So counting them, we get $11$ of the possible $36$ outcomes of which at least one $6$ is rolled. Now simply express this probability as a fraction!

http://www.wolframalpha.com/input/?i=chance+of+throwing+1+6%27s+with+2+dice

X = # of occurrences of 6 with 2 dice

P(X=1) = P(the first dice has 6, the second hasn't) + P(the second dice has 6, the first one hasn't) = 1/6 * 5/6 + 5/6 * 1/6 = 5/ 36 + 5/36 = 5/18 = 0.2778.

Which is also what Wolfram Alpha computes.

Then P(X=2) = 1/6 * 1/6 = 1/36.

5/18 + 1/36 = 11/36, which is indeed the answer.