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.
Consider resolution on a 2-SAT formula. Any resolvent is of size at most 2 (note that $n + m -2 \le 2$ if $n, m \le 2$ for clauses of length $n$ and $m$ resp). The number of clauses of size 2 is quadratic in the number of variables. Therefore, the resolution algorithm is in P.
Once you get to 3-SAT you can get bigger and bigger resolvents, so it all goes pear-shaped :).
Try translating a problem into 2-SAT. As you can't have clauses of size 3, you can't (in general) encode implications involving 3 variables or more, for instance that one variable is the result of a binary operation on two others. This is a huge restriction.
As Walter says, clauses of 2-SAT have a special form. This can be exploited to find solutions quickly.
There are actually several classes of SAT instances that can be decided in polynomial time, and 2-SAT is just one of these tractable classes. There are three broad kinds of reasons for tractability:
(Structural tractability) Any class of SAT instances where the variables interact in a tree-like fashion can be solved in polynomial time. The degree of the polynomial depends on the maximum width of instances in the class, where the width measures how far an instance is from being a tree. More precisely, Marx showed that if the instances have bounded submodular width, then the class can be decided in polynomial time using a divide-and-conquer approach.
(Language tractability) Any class of SAT instances where the pattern of true-false variables is "nice", can be solved in polynomial time. More precisely, the pattern of literals defines a language of relations, and Schaefer classified the six languages that lead to tractability, each with its own algorithm. 2-SAT forms one of the six Schaefer classes.
(Hybrid tractability) There are also some classes of instances that do not fall into the other two categories, but which can be solved in polynomial time for other reasons.
If you understand the algorithm for 2SAT, you already know why it's in P - this is precisely what the algorithm demonstrates. I think this comic illustrates my point. As you already know why 2SAT is in P, what you probably want to know is why 2SAT isn't NP-hard.
To understand why 2SAT isn't NP-hard, you have to consider how easy it is to reduce other problems in NP to it. To get an intuitive understanding of this, look at how SAT can be reduced to 3SAT and try to apply the same techniques to reduce SAT to 2SAT. 2SAT is just not as expressive as 3SAT and other SAT variants.