Concerning The Number of Ways of Drawing a Full House vs. Two Pair The Wikipedia entry for "Poker probability" gives the following result for the number of ways of drawing a full house:
$$ \binom{13}{1} \binom{4}{3} \binom{12}{1} \binom{4}{2} = 3, 744. $$
The logic here is conventional: From the 13 kinds you choose 1, then choose 3 of the 4 available cards of that kind. Then, from the remaining 12 kinds, you choose 1, then choose 2 of the available cards of that kind. No problem.
However, for the number of ways of drawing a two pair it gives the following:
$$ \binom{13}{2} \binom{4}{2} \binom{4}{2} \binom{11}{1} \binom{4}{1} = 123, 552. $$
But, by the logic used for drawing a full house, shouldn't it be the following:
$$ \binom{13}{1} \binom{4}{2} \binom{12}{1} \binom{4}{2} \binom{11}{1} \binom{4}{1} = 247, 104? $$
OR, conversely, by using the logic used to get a two pair, shouldn't the number of ways of drawing a full house be the following:
$$ \binom{13}{2} \binom{4}{3} \binom{4}{2} = 2,808? $$
As you can see, this isn't just a matter of inconsistent approaches that give the same result; these give different values. I just don't know how I'm supposed to be able to intuit these sort of things, seeing as they seem like the same thing to me.

In the second formula you give for the number of was to draw a two pair which is a product of $6$ terms, the outcomes (4, heart and spade, 3, heart and diamond, J, club) and (3, heart and diamond, 4, heart and spade, J, club) are counted as different hands when in fact they are the same five-card hand.  So you need to divide this value by $2$ to get the correct answer.  
In the second formula you give for the number of ways to draw a full house (your formula should evaluate to $1872$, not $2808$),  the first term ${13 \choose 2}$ is the number of ways to draw two kinds from $13$ kinds, and from these two kinds, you can choose the kind for the triple in $2$ ways (recall that a full house has both a triple and a pair).  So you need to multiply your answer $1872$ by $2$.  Here, since the order in which you draw the two kinds is important, using combinations (binomial coefficient) just once will not suffice.  
The reason you are using a different method for the two problems is that in a full house, the two subsets of cards have different sizes ($2$ and $3$) whereas if the hand is two pairs, then the two subsets of cards have the same size $2$. You are correct in thinking of the experiment as being a sequence of steps and taking the product of the number of ways to do each step, but during this process you need to be careful to not over-count or under-count.
