Okay, after some reading I think I understand the construction now. Start with the vorticity equation,$$\frac{D\boldsymbol{\omega}}{Dt} = (\boldsymbol{\omega} \cdot \nabla) \boldsymbol{u}$$where the ##D/Dt## is the material derivative. If the flow is confined to a plane, then ##\boldsymbol{\omega} = \nabla \times \boldsymbol{u}## is orthogonal to the plane, which contains ##\nabla \boldsymbol{u}##. That means, the RHS is zero, and $$\frac{D \boldsymbol{\omega}}{Dt} = \boldsymbol{0}$$Vortices with cylindrical symmetry are described by a velocity potential ##\phi = k\theta##, i.e. ##\boldsymbol{u} = \nabla \phi = (k/r) \boldsymbol{\theta}##. If we have N vortices, then the resulting velocity potential is$$\phi(\boldsymbol{r}) = \sum_i^N k_i\theta_i$$where the ##\theta_i## are the angles of the line segments from the ##i##th vortex to the point ##\boldsymbol{r}##, w.r.t. the ##x## axis. Helmholtz tells us that any given vortex only experiences the velocity field arising due to all of the other vortices (and not due to itself), so $$\dot{\boldsymbol{r}}_i = \nabla_{\boldsymbol{r}_i} \left( \sum_{j \neq i}^N k_j \theta_j \right) = \sum_{j \neq i}^N k_j \nabla_{\boldsymbol{r}_i} \theta_j = \sum_{j \neq i}^N k_j \left( \frac{\boldsymbol{z} \times (\boldsymbol{r}_i - \boldsymbol{r}_j)}{|\boldsymbol{r}_i - \boldsymbol{r}_j|^2} \right)$$and when we break that into ##x## and ##y## components, I think that agrees with what Prof. Tong wrote (except the ##\gamma## terms are wrapped inside the ##k## terms here).
