You're a bit confused about the terminology: in order to show that HAMPATH is NP-complete, you need to show that it's in NP (that's very easy) and that every problem in NP reduces to it. You do that by reducing some other NP-complete problem to HAMPATH.
As Theo noted, 3SAT is a special case of SAT, and so it's easier use 3SAT. The reduction that works for SAT will also work for 3SAT, but it may need to handle more cases. I'm not familiar with this particular reduction, but it might be that it's actually straightforward to generalize it to the "general case", but there's nothing to be gained by doing it.
Reductions showing a problem is NP-complete are usually like that - very intricate and slightly hard to follow. But usually there's the main idea - e.g., what corresponds to variables, what corresponds to clauses, why it should work - and the technical details. See first that you understand the main idea.
The technical details break into two parts. One part is the gadgets - you need to understand what the gadgets are supposed to do and how they accomplish it. The second part is proving that the construction works. In your case, it's probably easy to show that if the 3SAT instance is satisfiable, then the graph has a Hamiltonian path. It's probably more difficult to show that if the instance is unsatisfiable, the graph doesn't have any Hamiltonian paths. There's some intuition which leads us to believe that to be the case, but we still have to prove it, and the proof might be tricky.
Don't get discouraged if you don't "get it" right away. The best thing to understand these proofs is to do some of your own. Only then can you appreciate the thought processes behind such proofs. Better still, you might try working out a similar proof yourself, and see where you get stuck. There are some obstacles to overcome, and if you're aware of them then the construction will become much more interesting.
Looking at the reduction, I think Sisper explains it pretty well. In fact, for an expert it might be enough to mostly look at all the figures. The pattern of this reduction is very standard, so once you understand this one, you'll be in much better position to understand a whole lot of them!
The idea is to have variable-gadgets and clause-gadgets. The variable-gadgets can be traversed in two ways (according to the truth assignment). The clause-gadgets can be traversed-through from a satisfying literal.
So far for the idea. The construction itself is pretty simple, and it's easy to check that if a formula is satisfiable then there's a Hamiltonian path. It's also easy to see that if there's a similarly-looking Hamiltonian path then there's a satisfying assignment. The hard part is showing that all Hamiltonian paths "look as if they came out from a truth assignment", which must be satisfying.
An important part of these gadget schemes is that there is a 3-clique somewhere called the palette that arbitrates the logic roles of colors. In other words, there are 3 nodes all connected to each other, labeled $v_T$, $v_F$, and $v_R$. The palette can be arbitrarily colored, and whatever colors are chosen represent the 3 logic values. Then when a gadget has a fixed "true" node in it, arcs are extended from fixed $v_F$ and $v_R$ nodes to it, to compel its color to be the same as $v_T$.
So the reduction doesn't require any strange modifications to the concept of the 3-COL problem. The reduction algorithm generates a planar graph, and that graph can be 3-colored iff the 3-SAT problem can be satisified.
Best Answer
For the pedantic's sake, we first have a polynomial reduction $3SAT \leq_p s3SAT$, where the later has strictly 3 terms per clause, not less (as accepted by the former). This is achieved by taking an instance of $3SAT$ and mapping the clauses respectively:
$ (a) \mapsto (\lnot s \lor \lnot s \lor \lnot s) \land (s \lor s \lor a) $
$ (a \lor b) \mapsto (\lnot s \lor \lnot s \lor \lnot s) \land (s \lor b \lor a) $
where $s$ is a dummy variable, and a and b are terms where $\lnot \lnot$ is negated respectively. We concatenate the end result with $\land$. Clearly we construct in polynomial time.
Next we do the polynomial reduction $s3SAT \leq_p NAE-4SAT$, where the second is simply $NAE-3SAT$ but with four terms per clause. For this we take an instance of $s3SAT$ and map each clause as follows:
$ (x \lor y \lor z) \mapsto (x \lor y \lor z \lor s) \land (\lnot x \lor \lnot y \lor \lnot z \lor \lnot s)$
Where $x,y,z$ are terms and $s$ is our dummy variable, as before. Notice the symmetry here, if we find an assignment with $s = true$ we can simply invert the assignment to receive another valid assignment with $s=false$. This is the assignment we want.
As a last step we reduce $NAE-4SAT \leq_p NAE-3SAT$. Take an instance in $NAE-4SAT$ and map the clauses as follows:
$ (a \lor b \lor c \lor d) \mapsto (s \lor a \lor b) \land (\lnot s \lor c \lor d)$
All the same as before, concatenate the result once more. Notice here that if the true and false variable's (one of each must exist) are mapped to the same clause, $s$ can be choose appropriately. If they are mapped to different clauses, $s$ can be choose opposite to the respective variable value in each pair.
To summarize: $3SAT \leq_p s3-SAT \leq_p NAE_4SAT \leq_p NAE-3SAT$.
Answer to extra question: 3SAT only allows $\lor$ in the clauses.