I thought the whole point of BOE was to separate electronic and nuclear equations, but here they seem to integrate the nuclei-nuclei interaction in the electronic term (they only leave the nuclear kinetic energy out).
We're working with 2 sets of coordinates: $\mathbf{r}$, for the electrons, and $\mathbf R$ for the nuclei. In the electronic term, we've gone and the "nailed" down the nuclei, in some predetermined positions, letting the electrons move freely. The term itself doesn't interact with the electrons, and is meant to represent the potential energy contribution of the static configuration of the nuclei. It doesn't really matter if you include it or not: it's only worth a constant at the end of the day.
In my understanding of what is above in the derivation, I thought that the very definition of $(H_n(\boldsymbol{R}))_{k'k}$ was: $(H_n(\boldsymbol{R}))_{k'k} = < \chi_k'|T_n|\chi_k>_{(\boldsymbol{r})}.$
Wikipedia is being very confusing with it's notation. To see why $(H_n(\boldsymbol{R}))_{k'k} = < \chi_k'|T_n|\chi_k>_{(\boldsymbol{r})}$ is wrong, lets just apply $\hat H$ to $\Psi(\mathbf r, \mathbf R)$:
$${\displaystyle H\Psi (\mathbf {R} ,\mathbf {r} )=E\Psi (\mathbf {R} ,\mathbf {r} )}, $$hence, $$\left[H_{\text{e}}+T_{\text{n}}\right]\sum _{k=1}^{K}\chi _{k}(\mathbf {r} ;\mathbf {R} )\phi _{k}(\mathbf {R} ) = E\sum _{k=1}^{K}\chi _{k}(\mathbf {r} ;\mathbf {R} )\phi _{k}(\mathbf {R} ).$$
Hitting the right side with $\chi _{k'}^*(\mathbf {r} ;\mathbf {R} )$, and integrating over the electronic wavefunctions (notice the $*$ and $k'$ I've attached!),
$$\int d\mathbf r \Big(\chi _{k'}^*(\mathbf {r} ;\mathbf {R} )\left[H_{\text{e}}+T_{\text{n}}\right]\sum _{k=1}^{K}\chi _{k}(\mathbf {r} ;\mathbf {R} )\phi _{k}(\mathbf {R} )\Big) = \int d\mathbf r \Big(\chi _{k'}^*(\mathbf {r} ;\mathbf {R} ) E\sum _{k=1}^{K}\chi _{k}(\mathbf {r} ;\mathbf {R} )\phi _{k}(\mathbf {R} )\Big).$$
On the right side, we may invoke the orthonormality of $\chi_k$ since $E$ is a constant, but on the left side, $H_e$ and $T_n$ are operators, and we can't. Fortunately, we can still separate left side and simplify it considerably, like this:
$$\begin{align*}&\int d\mathbf r \Big(\chi _{k'}^*(\mathbf {r} ;\mathbf {R} )\left[H_{\text{e}}+T_{\text{n}}\right]\sum _{k=1}^{K}\chi _{k}(\mathbf {r} ;\mathbf {R} )\phi _{k}(\mathbf {R} )\Big) \\
= &\int d\mathbf r \Big(\sum _{k=1}^{K}\chi _{k'}^*(\mathbf {r} ;\mathbf {R} )H_{\text{e}}\chi _{k}(\mathbf {r} ;\mathbf {R} )\phi _{k}(\mathbf {R} ) + \sum _{k=1}^{K}\chi _{k'}^*(\mathbf {r} ;\mathbf {R} )T_{\text{n}}\chi _{k}(\mathbf {r} ;\mathbf {R} )\phi _{k}(\mathbf {R} )\Big) \\
= & E_e(\mathbf{R})\phi _{k'}(\mathbf {R} ) + \sum _{k=1}^{K}\int d\mathbf r \Big(\chi _{k'}^*(\mathbf {r} ;\mathbf {R} )T_{\text{n}}\chi _{k}(\mathbf {r} ;\mathbf {R} ) \phi _{k}(\mathbf {R} )\Big)\end{align*}.$$
These represent a total of $K$ equations. The first term is the same as the Wikipedia term, just in component form. The second term looks somewhat familiar if we choose to define
$$\int d\mathbf r \chi _{k'}^*(\mathbf {r} ;\mathbf {R} )T_{\text{n}}\chi _{k}(\mathbf {r} ;\mathbf {R} ) = \big (\mathbb {H} _{\text{n}}(\mathbf {R} ){\big )}_{k'k},$$
but it's clear that this isn't the case, because $T_{\text{n}}$ is really acting on $\chi _{k}(\mathbf {r} ;\mathbf {R} ) \phi _{k}(\mathbf {R} )$ inside the integral, not just $\chi _{k}(\mathbf {r} ;\mathbf {R} )$. It certainty helps to use the matrix notation to clean up the clutter, but it's easy to forget what is really going on if you're not famailiar with it. If you use the product rule to expand out $T_{\text{n}}\chi _{k}(\mathbf {r} ;\mathbf {R} ) \phi _{k}(\mathbf {R} )$, it will lead to the relation you want. If you find anything confusing, feel free to ask.
