I've come across the polynomial algorithm that solves 2SAT. I've found it boggling that 2SAT is in P where all (or many others) of the SAT instances are NP-Complete. What makes this problem different? What makes it so easy (NL-Complete - even easier than P)? 
Here is a further intuitive and unpretentious explanation along the lines of MGwynne's answer.

With $2$-SAT, you can only express implications of the form $a \Rightarrow b$, where $a$ and $b$ are literals. More precisely, every $2$-clause $l_1 \lor l_2$ can be understood as a pair of implications: $\lnot l_1 \Rightarrow l_2$ and $\lnot l_2 \Rightarrow l_1$. If you set $a$ to true, $b$ must be true as well. If you set $b$ to false, $a$ must be false as well. Such implications are straightforward: there is no choice, you have only $1$ possibility, there is no room for case-multiplication. You can just follow every possible implication chain, and see if you ever derive both $\lnot l$ from $l$ and $l$ from $\lnot l$: if you do for some $l$, then the 2-SAT formula is unsatisfiable, otherwise it is satisfiable. It is the case that the number of possible implication chains is polynomially bounded in the size of the input formula.

With $3$-SAT, you can express implications of the form $a \Rightarrow b \lor c$, where $a$, $b$ and $c$ are literals. Now you are in trouble: if you set $a$ to true, then either $b$ or $c$ must be true, but which one? You have to make a choice: you have 2 possibilities. Here is where case-multiplication becomes possible, and where the combinatorial explosion arises.

In other words, $3$-SAT is able to express the presence of more than one possibility, while $2$-SAT doesn't have such ability. It is precisely such presence of more than one possibility ($2$ possibilities in case of $3$-SAT, $k-1$ possibilities in case of $k$-SAT) that causes the typical combinatorial explosion of NP-complete problems. 
