indeed reveals a contribution to the change in total energy due to the change of the wave functions $\Phi_j$.
However, provided that the $\Phi_j$ are eigenstates of $H$, it is easy to show that the last two terms cancel each other and in the special case of $H=T+V$ the force is given by
\begin{equation}
-\vec{F}_i=-\sum_j \Phi_j^*\Phi_j\frac{\partial V}{\partial \vec{R}_i}
+\vec{F}_i=-\sum_j \langle \Phi_j | \Phi_j\frac{\partial V}{\partial \vec{R}_i} \rangle
\text{ .}
\end{equation}
This is called the Hellmann-Feynman theorem \cite{feynman39}, which enables the calculation of forces, called the Hellmann-Feynman forces, acting on the nuclei for a given configuration, without the need for evaluating computationally costly energy maps.