I suppose you are in the context of a rotating black hole, which is described by the Kerr metric, which in Boyer–Lindquist coordinates is ($c = 1$)
\begin{equation*}
ds^{2} = - \left( 1 - \frac{GMr}{\rho^{2}} \right) dt^{2} - \frac{4GMar \sin^{2}\theta^{2}}{\rho^{2}} dt d\varphi + \frac{\rho^{2}}{\Delta} dr^{2} + \rho^{2} d\theta^{2} + \frac{(r^{2} + a^{2})^{2} - a^{2} \Delta \sin^{2}\theta}{\rho^{2}} \sin^{2}\theta d\varphi^{2},
\end{equation*}
where $\rho^{2} = r^{2} + a^{2} \cos^{2}\theta, \Delta = r^{2} - 2GMr - a^{2}$ and $M, a$ are the "mass" and "angular momentum" of the black hole, respectively.
Energy is the dot product between the four momentum of the object and the four-velocity of the observer
The energy you are referring to is the energy of a particle (or a light ray) whose trajectory is given by its four-momentum $p^{\mu}$, measured by an observer whose trajectory is given by its four-velocity $u_{obs}^{\mu}$. That is,
\begin{equation*}
E_{\text{obs}} = -p_{\mu} u_{\text{obs}}^{\mu},
\end{equation*}
where we add the minus sign simply because is the covariant version of the energy measured by an observer in special relativity (SR). This energy is, in principle, different from conserved mathematical invariant due to $\partial_{t}$ being a Killing vector field, which is given by
\begin{equation*}
E = - g(\partial_{t}, u) = - g_{\mu \nu} \delta^{\mu}_{t} u^{\nu} = - g_{t \nu} u^{\nu} = - u_{t},
\end{equation*}
where $u^{\mu}$ is the four-velocity of the particle (or light ray), which can be interpreted as an energy measured by an observer at infinity. Note that we also add the minus sign for the same reason (and ultimately, since $\partial_{t}$ is time-like). In the Penrose process, one must consider the latter energy notion, and the reason it works (theoretically at least) is because of the sign game of this invariant. Precisely, the magnitude $E$ is always positive outside the ergoregion, and can be either positive or negative inside the ergoregion. This is easy to see, since there can't be a particle with $E = 0$ outside the ergoregion (if the particle ever gets far enough, SR should be valid, where $E_{obs}$, which would then coincide with the invariant $E$, must be at least $m$). Conversely, since the hypersurface $\{ g_{tt} = 0 \}$ is time-like, particles can go through it, and since there clearly exist particles which eventually fall to the black hole, the energy $E$ inside the ergoregion can be positive. In order to see that it can also be negative, consider the concrete expression of $E$,
\begin{equation*}
E = - g_{t \nu} u^{\nu} = - g_{t t} u^{t} - g_{t \varphi} u^{\varphi} = \left( 1 - \frac{GMr}{\rho^{2}} \right) \dot{t} + \frac{2GMar \sin^{2}\theta^{2}}{\rho^{2}} \dot{\varphi},
\end{equation*}
and simply note that if a particle ever passes through $\theta = 0$, then its energy $E$ must be negative, since $g_{tt}$ is positive inside the ergosphere.
If the notion of negative energy is bugging you (which is good), it is important to realize that $E$ cannot be always interpreted as energy, but solely a mathematical invariant that regains physical value at infinity.
when the object rotates in the same direction as the black hole, it has positive energy. If the object rotates in the opposite direction as the black hole, it has negative energy
Inside the ergoregion, objects cannot rotate in the opposite direction as the black hole, and furthermore, they must rotate in the same direction, so this would not make any difference in the Penrose process.
