For an isomorphism of sheaf on X $f : \mathscr{F} \to \mathscr{G}$, suppose it is "locally isomorphic," that is, there is an open cover $U_i$ such that $f|U_i: \mathscr{F}_{U_i} \to \mathscr{G}_{U_i}$ is an isomorphism for all $U_i$.
I cannot specify where is wrong about the argument below :
$f$ is isomorphic
iff $f_x$ is isomorphic for all $x\in X$
iff $f_x$ is isomorphic for all $U_i$ and $x\in U_i$
iff $f|U_i$ is isomorphic for all $U_i$.
Your equivalences you're listing are correct. The isomorphisms already glue since they given by $f$. For example, the correct statement for the second point is "There exist a morphism $f : \mathscr F \to \mathscr G$ such that for all $x \in X$, $f_x$ is an isomorphism."
The wrong statement would be to assume that if $f_x : \mathscr F_x \to \mathscr G_x$ is an arbitrary family family of isomorphisms, then $\mathscr F \cong \mathscr G$ are isomorphic (e.g a counter-example is given by any non-trivial locally free sheaf).