BIG-Oh means :
if there are two functions f(n) and g(n) such that
f(n) <= C.g(n) where exists two constants C and n0 such that C>0 and n>=$n_{0}$
then f(n)= O(g(n))
that means if using a constant C we can prove f(n)<=C.g(n) for any value of n>= $n_{0}$ then we can say f(n)=O(g(n)).
so for your question if f(n)=2n and g(n)=n then if we can prove f(n)<=C.g(n) for all n where n>= constant $n_{0}$, then we can say 2n=O(n).
to prove that let constant C=3 then
f(n)<=3 g(n) from all n greater or equal to $n_{0}$ = 1 onwards
2n <= 3n from all n greater or equal to $n_{0}$ = 1 onwards. hence it is proved that 2n=O(n).
now, $2^{f(n)}$ = $2^{2n}$ and $2^{g(n)}$ = $2^{n}$
here we can see f(n)> g(n) by $2^{n}$ times hence using any constant we can't prove that $2^{f(n)}$ = O($2^{g(n)}$)
so if, f(n)= O(g(n)) then $2^{f(n)}$ = O($2^{g(n)}$) need not be true always.
i hope your doubt is cleared now, Thankyou.
