Does polynomial time reduction from CNFFAL to CNFSAT is also polynomial time reduction from CNFSAT to CNFFAL?
No, $CNFFAL\le_P CNFSAT$ doesn't appear to imply $CNFSAT\le_P CNFFAL$, as far as we know.   

As you say, we know that $CNFFAL\le_P CNFSAT$ is true.  However, $CNFSAT\le_P CNFFAL$ is true if and only if $P = NP$.

Consequently, the claim $CNFFAL\le_PCNFSAT\implies CNFSAT\le_PCNFFAL$ is logically equivalent to the claim that $P=NP$.  There is no known proof that $P=NP$ (and many would consider it surprising if that is true).

I didn't understand why you think the implication does hold.  I couldn't quite follow your reasoning.  The notation $\beta$ is never explained clearly (I don't know what you mean by "the boolean formulated generated by a NTM"; NTMs don't generate boolean formulas, so that would need to be defined more clearly). It's hard to know what you might be thinking, but if I were forced to speculated, I'd guess that the flaw in your reasoning might have something to do with the situation where $\alpha$ is both satisfiable and falsifiable.  Then it's true that $NTM_{CNFFAL}$ has both an accept path and a reject route, but so what?  It's not clear why that would imply anything interesting.