Why is Boolean satisfiability such a rare case? In the space of all K-sat formulas, True and False should have an equal set size. For every un-Satisfiable formula (F), there will an F' (or F-prime) which will be Satisfiable by definition. I cannot get my head around this. 

Edit: Let me put it this way: I have constructed a circuit based on a Boolean formula, it is not Satisfiable, now I add one NOT-gate just before the output of the circuit. Does it make sense?
Your trick doesn't really work.  There are several issues.

First, your trick shows that for every unsatisfiable circuit of size $n$, there exists a satisfiable circuit of size $n+1$.  But that doesn't tell you anything about the relationship between the number of unsatisfiable circuits of size $n$ vs the number of satisfiable circuits of size $n$.  And, in general, there are way more circuits of size $n+1$ than circuits of size $n$, and most circuits are satisfiable, so the mapping you've found doesn't prove what you want it to prove.  To prove what you're trying to prove, you'd need to find a bijection between unsatisfiable circuits of size $n$ and satisfiable circuits of size $n$ (not $n+1$); no such bijection exists.

Second, the mapping you describe isn't a bijection.  Consider for example the circuit $\phi(x) = x$.  This is satisfiable.  Slapping a "not" in front if it, to get $\phi'(x) = \neg x$, yields another circuit that is also satisfiable.  Many circuits are both satisfiable (there exists an assignment that makes $\phi$ true) and their-complement-is-satisfiable (there exists a different assignment that makes $\phi$ false).  So negating an unsatisfiable circuit makes it satisfiable, but negating a satisfiable circuit doesn't necessarily make it unsatisfiable.

Finally, there's the distinction between circuits vs formulas, as David Richerby explains.  If you negate the output of a CNF formula, the result isn't a CNF formula.