There might be 21 possible outcomes ( (1,1), (1,2),....(6,6) ) , but they don't have equal probability, 1+1 can only occur in one way, but 2+3 can occur two different ways. There are 6 ways to throw (1,1), (2,2), (3,3), etc, and 15 different pairs (1,2)/(2,1), (1,3)/(3,1), etc, making 36 possible combinations. We can get a 2 and 3 in two different ways, so the probability is 2/36 as you say.
When doing these kinds of computations in probability, it's often easier to consider the opposite event of what you're actually interested in. Here, the event you care about is "at least one die is a 6" (for example), so the opposite event is "no die is a 6." This means that die 1 is not a 6, and die 2 is not a 6, and dice 3, 4, 5 are all not a 6. The key in this computation is the word "and," because the independence of the dice means we can multiply those individual probabilities. The chance that no die is a 6 is therefore
$$\frac 5 6 \cdot \frac 5 6 \cdot \frac 5 6 \cdot \frac 5 6 \cdot \frac 5 6 = \frac{3125}{7776}$$
and the probability of the event you want is thus $ 1 - \frac{3125}{7776} = \frac{4651}{7776} \approx 0.5981$.
To answer your question a bit more comprehensively, you asked about when to use the addition rule and when to use the multiplication rule. At a basic level, multiplication corresponds to the word "and," and addition corresponds to the word "or." However, there are caveats; multiplication corresponds to "and" when events are independent, and addition corresponds to "or" when events are disjoint. Recognizing when these things happen is one of the main challenges of learning probability theory and comes only with a fair bit of experience in the subject, as far as I can tell.
In this case, the reason I knew to consider the complementary event was that the main event you wanted to consider had to be expressed in terms of several "or" events that were relatively complicated (i.e. "exactly 1 die is a 6," "exactly 2 dice are a 6," etc.) but the opposite event was not complicated in that way.
Best Answer
You have computed the probability that (for example) the first and second throws of the pair of dice produce pairs (doublets) and the others do not.
It could also be the case that the first throw is a pair, the second is not, but the third throw is (and the remaining two are not). This is also an event where you get a pair exactly twice, but it is disjoint from the other event.
So you must also consider how many possible different events you could be looking for. The two pairs could come anywhere within the sequence of throws.