What is $v \, dp$ work and when do I use it?

$PdV$ is boundary work. $VdP$ is isentropic shaft work in pumps (as you have identified above), gas turbines, etc. Now you must realize that even in a pump or turbine the mechanism of work is still $Pdv$, i.e., the gas pushing on the blade out of its way. But, then there is work required to maintain the flow in and out of the device/control volume, which requires flow work $PV$ so the net reversible work from a steady-flow device turns out to be shaft $vdP$.

Why flow work $PV$? To push a packet of fluid with volume $V$ forward into a device you have to do work against the pressure of the fluid already in the device, i.e., overcome the back force of that fluid. This implies the work you do in pushing your new packet of length $L$ and cross-section area $A$ into the device is: \begin{align*} \int Fdx = \int_{0}^{L}PAdx = PV \end{align*} It must be noted that in a steady-flow device (unlike in a piston) the back pressure $P$ is constant.

Now consider the device (e.g., turbine to be a control volume). The energy of the fluid going in is its internal energy and the work invested into the fluid to enter the device: $U_{entry}+P_{entry}V_{entry}=H_{entry}$. Similarly for exit from the device. The net change across the device is $\Delta H$. For a differential device (or across a small change) this is $dH$. The work output from the shaft of then device is the $\delta W= dH$.

Now if the device is isentropic, i.e., adiabatic-reversible. The Gibbs equation provides: \begin{align*} &dH=TdS+VdP=VdP\\ &\delta W =dH=VdP \qquad (\text{isentropic}\; dS=0) \end{align*}

Therefore $VdP$ is isentropic shaft work from a flowing device.

Important points: 1) Both internal energy and enthalpy are state variables, therefore can be measured for a system static or flowing. This is why sometimes there is a tendency to use $U$ and $H$ incorrectly. The true purpose of $H$ is to capture the work required to push/maintain a flow against a back pressure, i.e., it incorporates the $PV$ part. Therefore when you write an energy balance with flows coming in and out, the energy crossing boundary is not just $U$ but $H$ and this distinction must be kept in mind.

2) $VdP$ is isentropic steady-flow shaft work. The isentropic is key here.

$\delta W=\mathrm{d}(PV)$ is wrong. We always have $\delta W=P\mathrm{d}V$ (unless there are other interactions like magnetic field). In fact, the reason that many books opt to denote infinitesimal work as $\delta W$ instead of $\mathrm{d}W$ is to emphasize that work is not an exact differential.

$V\mathrm{d}P$ manifests in these formulas $$\mathrm{d}H=T\mathrm{d}S+V\mathrm{d}P$$ $$\mathrm{d}G=-S\mathrm{d}T+V\mathrm{d}P$$ where $H$ and $G$ are not internal energy, but enthalpy and Gibbs free energy. These two concepts are more useful than internal energy in isobaric processes.

Sankaran is correct in that the magnitude of the net reversible shaft work $\delta w$ is $vdP$, but he is incorrect in ascribing to this quantity a positive sign. In reality, the shaft work done by the system must be $-vdP $, here's why...

If, by definition, a differential change in enthalpy $dh = TdS + vdP$, then for an adiabatic process where $\delta q = TdS = 0$, the differential change in enthalpy must be

$dh = (0) + vdP = vdP$.

Now, if the work done by the system is positive ($\delta w > 0)$ its change in enthalpy must be negative $(dh < 0)$. These two facts in conjunction must mean that the net reversible shaft work done by the system is:

$\delta w = -vdP = -dh$

I ran into this problem myself in trying to derive the specific work produced by an isentropic turbine, and you will definitely get the wrong answer if you start with the premise that the differential shaft work is simply $vdP$.