Why do they call it decision problems, if it's really about finding a solution that is much more than a boolean (like a path, ...)?
The whole idea of "verify" versus "find" makes no sense for problems with a yes/no answer, because if you can verify a boolean answer in polynomial time, you can always find the solution in polynomial time too, after all there are only two possible solutions so you need to verify only one to find the answer.
If you need extra data to prove this "yes" in polynomial time, then the answer isn't a boolean anymore, but a boolean plus the extra data, so how can you still call that a yes/no problem?
...if you can verify a boolean answer in polynomial time, you can always find the solution in polynomial time too, after all there are only two possible solutions so you need to verify only one to find the answer.
It doesn't work that way. The common characteristic of NP-complete problems is that there are an exponential number of possible solutions, but verifying that a purported solution is a solution takes only polynomial time.
For example consider SAT (see http://en.wikipedia.org/wiki/Boolean_satisfiability_problem for more). This is the problem of deciding whether there is a set of possible variable values that make a given Boolean equation evaluate to True. For an equation of length n there are in general O(n) variables, and therefore 2^O(n) possible sets of true/false values that the variables can have. So a brute force search takes exponential time.
But if you give me an equation, AND a set of values for the variables that you claim will result in the final expression evaluating to True, I can verify it in time O(n).
"If you need extra data to prove this "yes" in polynomial time, then the answer isn't a boolean anymore, but a boolean plus the extra data, so how can you still call that a yes/no problem?"
You are combining two distinct concepts: the answer and the 'evidence'. The answer to an NP-complete problem is always Yes or No. But, we say that we can "verify" a decision problem in polynomial time if, given an answer AND evidence, we can check that the answer is true.
For concreteness: the k-clique problem (determining if there's a clique of at least size k in a graph) is hard. But, if somebody told us the answer was Yes, and gave us appropriate evidence (say, a set of k nodes in the graph which form a clique), verifying would be easy.
Just to give a little more context, because I think this gripe is very common:
There exists a proof that NAND gates are all you ever need to compute an arbitrary boolean function from M bits to N bits. Actually, you also need a "copy" gate -- the ability to generate two bits with value X from one bit with value X -- and "wires" to communicate values from gate to gate, but that's about it.
The proof starts out like this: "a function from M bits to N bits is actually just N functions from M bits to 1 bit, as long as I can copy the input N times and feed it to the N functions which compute each output bit. So if I solve the problem for M-to-1, I automatically solve M-to-N as long as I have COPY and SWAP gates."
It's very important to see that the exact same approach is being used in the P-vs-NP question. P and NP are designed to be about functions which take M bits to 1 bit, hence they are "decision problems" with a boolean output but a variable input.
The question "P = NP?" can also be phrased more common-sensically, if not totally precisely, as "is brute force ever really necessary?". What is brute force? It is if you have to try a substantial fraction of the 2^M inputs to find a certain 1-bit output. What does this look like here? Well, that gets more complicated because we use scaling of a problem and a system of solution to discuss what's going on.
So a problem is in NP if (1) you can reduce it to some sort of M-bit input to 1-bit output, and (2) there exists some constant number K such that if I double the input size from M to 2M, the circuit has no more than K times the number of gates. This gives the logic circuit a "polynomial-time" size.
Then a problem is in P if we can find an input which makes the output 1 in a polynomial number of gates as well. Thus this is the "inverse" problem, "find me an input which makes this function 1." In fact we've just defined what's called a "satisfiability problem" -- find an input which satisfies this polynomial-size M-variable boolean formula.
A problem is in NP-complete if the recursion holds: if you can implement COPY and NAND and wires in that system in a polynomial number of gates, so that any problem in NP can be written in terms of your problem. This is part of the value of having these "constant K" constraints rather than specifying a given K; we get to define these NP-complete cases as the "hardest cases in NP," because if you can reverse them, then you can reverse any other problem in NP.
The decision problem is formulated something like "Is there a hamiltonian path of at most length n" for different n. You can get the actual answer by polling the algorithm that answers the decision problem using a binary search plugging in different values for n (since there is a maximum number of possible paths through a graph with a certain number of nodes)
The whole idea of "verify" versus "find" makes no sense for problems with a yes/no answer, because if you can verify a boolean answer in polynomial time, you can always find the solution in polynomial time too, after all there are only two possible solutions so you need to verify only one to find the answer.
If you need extra data to prove this "yes" in polynomial time, then the answer isn't a boolean anymore, but a boolean plus the extra data, so how can you still call that a yes/no problem?