Here the $\nabla_k$ denote that the gradient must be calculated at the point $r_k$.
Suppose I have a scalar function $V = V(\textbf{r}_1,\textbf{r}_2,\textbf{r}_3)$, where $\textbf{r}_k$ are 3 dimensional vectors defined as $\textbf{r}_k = \textbf{r}_k(x,y,z)$
Now I want to know how this symbol looks like: $\nabla_k$
I think it is like this : $\nabla_k V = \frac{\partial V}{\partial \textbf{r}_k} = \left(\frac{\partial V}{\partial r_{kx}},\frac{\partial V}{\partial r_{ky}},\frac{\partial V}{\partial r_{kz}}\right)$
For the denominator, $\partial r_{kx}$ is the $x$ variable of the kth vector $\textbf{r}_k$ (same for $y$ and $z$).
For an example, if we let $\textbf{r}_1 = \textbf{r}_1(x_1,y_1,z_1)$, where $x_1,y_1,z_1$ are variables of the $\textbf{r}_1$ vector. Then $\nabla_1 V = \frac{\partial V}{\partial \textbf{r}_1}= \left(\frac{\partial V}{\partial x_1},\frac{\partial V}{\partial y_1},\frac{\partial V}{\partial z_1}\right)$?
this cite offers more information: here for equations: 28 and 29.
I understand that this looks like a physics problem, but the notation is mathematical.
This is my expansion of the above(if correct which seems to hold but not sure.):
Where we go on to define $V$ which is in terms of generalised coordinates $q_i$ instead of $\textbf{r}_k$, thus we can replace the final $\frac{d \tilde V}{d q_i}$ with $\frac{\partial V}{\partial q_i}$ and arrive at the same conclusion as the text shows.