 Solution

A circular orbit has uniform angular motion, so its Keplerian radial-velocity curve is sinusoidal. Choose phase $\varphi =(t-t_{\mathrm{tr}})/P$, positive stellar velocity away from the observer, and an edge-on circular orbit. One consistent convention isTransit occurs at the decreasing systemic-velocity crossing $\varphi =0$; secondary eclipse is at the increasing crossing $\varphi =1/2$. Extrema occur one quarter of an orbit away. Reversing the velocity sign convention reverses the crossing directions but not the conjunction phases.

For an eccentric orbit, $\nu$ is not proportional to time. Solve $\mathcal{M}=E-e\sin E$, then use $\tan (\nu /2)=\sqrt{(1+e)/(1-e)}\tan (E/2)$ in the stated velocity relation. Motion is fastest near periastron, so the curve has an abrupt feature there and a more prolonged change near apastron, with its shape also depending on $\varpi$. It is not merely a shifted sinusoid. An eccentric transit need not coincide with $V=V_0$, because the $e\cos \varpi$ offset remains at conjunction.