Why don't we consider VΔP when we define Q?
Let's start by the fundamental ideas. Assume, you have a container with volume $V$ with a piston on it of  cross sectional area $A$ and container is filled with gas.
Suppose, you push the container by $\mathrm dx$ amount. So, the gas is compressed and work is done on the gas.
By the definition of work classically,
$$\mathrm dW=F\,\mathrm dx$$
Now if we use the fact that $P=\frac FA$, and $\mathrm dV=A\,\mathrm dx$, we will have $$\mathrm dW=\frac FAA\,\mathrm dx= P\,\mathrm dV$$ So, the classical definition of work only tells us the that work done on the gas should be $$\int_{V_1}^{V_2}P\,\mathrm dV=P\,\Delta V$$ if $P$ is constant.
So, the definition tells us there is no $V\,\Delta P$ term in the expresion for work. And, therfore, in the first law of thermodynamics, i.e. $\Delta Q = \Delta U - W_\text{on the system}$, also, that expression doesn't occur from work, and it is not also included in $\Delta U$ either.
But if you consider enthalpy change for a system, there $V\,\mathrm dP$ term will occur as $$\mathrm dH=\mathrm dU+\mathrm d(PV)=\mathrm dU+P\,\mathrm dV+V\,\mathrm dP$$