I'm trying to give a partial answer.
(ii) $\Rightarrow$ (i) is trivial.
(i) $\Rightarrow$ (ii). I cannot prove it completely, but the following argument should apply to the case $f(t,x)=g(t)h(x)$, where the variables are separable.
Assume (i). $M_t:=h(X_t) − h(X_0) −\int_0^t Ah(X_s)ds$ is a continuous martingale, with $M_0=0$. Apply the integration by parts formula for stochastic integrals, \int_0^t g(s)dM_s=g(t)M_t-\int_0^tM_sg'(s)ds. The left-hand side is a martingale. One can show that the right-hand side equals g(t)h(X_t)-g(0)h(X_0)-\int_0^t [g'(s)h(X_s)+g(s)Ah(X_s)]ds, because \int_0^tg'(s)\int_0^s Ah(X_u)duds=g(t)\int_0^tAh(X_s)ds-\int_0^tg(s)Ah(X_s)ds. Thus, the spacial case is proved. To get the general result, use some kind of extension argument?