Yes, you do need the integrand $f(X_t)$ to be predictable.  If it is merely adapted you may not get a martingale.

Intuitively, the stochastic integral $\int Y_t\,dM_t$ tells you the profit from a stock trading strategy.  $M_t$ is the share price at time $t$, and $Y_t$ is the number of shares you hold at time $t$.   Requiring that $Y$ be predictable means that in determining your holdings at time $t$, you can only use information about the stock prior to time $t$.  If $Y$ is adapted, you may also use the stock price at time $t$.  But since our compensated Poisson process $M_t$ moves by jumping up and drifting down, if $Y$ is adapted, you can buy shares at the instant the price jumps up to turn a quick profit, and sell them a short time later to lock it in.  That strategy gives you a guaranteed profit, so it can't be a martingale.

Formally, consider a time-homogeneous Poisson process ($\lambda(t) = 1$) and let $T$ be the time of the first jump: $T = \inf\{t : M_t > M_{t-}\}$.  $T$ is a stopping time, so if we let $X_t = 1_{\{T \le t < T+\epsilon\}}$ where $\epsilon < 1$, we can check that $X_t$ is adapted (but not predictable).  Thus our strategy is "buy 1 share when the stock jumps for the first time, sell it $\epsilon$ seconds later".

Let $I_t = \int_0^t X_t\,dM_t$ be the stochastic integral.  You can check that $I_t \ge 0$, and for large $t$ we have $I_t \ge 1-\epsilon$ with high probability, since it's likely there has been a jump by time $t$.  $I_t$ could be even larger if $M_t$ made additional jumps in the interval $[T, T+\epsilon)$.  But in any case, $I_t$ certainly is not a martingale.