The first thing you need to do is stop thinking about heat capacity in terms of heat Q. In thermodynamics, we define heat capacity in terms of internal energy U and enthalpy H, not in terms of heat. In this way, heat capacity is a physical property of the material being processed, and not a function of the process path. So once you get rid of all the equations in your post that involve Q, things get much simpler. For an arbitrary material (liquid, solid, or gas), the two heat capacities are defined as follows:
$$C_v=\left(\frac{\partial U}{\partial T}\right)_V$$
$$C_p=\left(\frac{\partial H}{\partial T}\right)_P$$
Secondly, for an ideal gas, the internal energy and enthalpy are functions only of temperature, and do not depend on volume or pressure. So, for an ideal gas, irrespective of process path, we always have that $$dU=C_vdT$$ and $$dH=C_pdT$$irrespective of whether the volume or pressure are constant.
Thirdly, please note that the heat transferred in a constant pressure process for an ideal gas is not equal to the heat transferred in a constant volume process for an ideal gas. In a constant volume process (assuming temperature independent heat capacities), $$\Delta U=C_v\Delta T=Q$$ and $$\Delta H=C_p\Delta T=\Delta U+V\Delta P=C_v\Delta T+V\Delta P=C_v\Delta T+R\Delta T=Q+R\Delta T$$
In a constant pressure process, $$\Delta U=C_v\Delta T=Q-P\Delta V=Q-R\Delta T$$and $$\Delta H=\Delta U+P\Delta V=C_v\Delta T+R\Delta T=\Delta U+R\Delta T=Q$$
So the change in enthalpy is not equal to the change in internal energy for either of these processes (as expected, since they are different functions of temperature). The only way that $\Delta H$ and be equal to $\Delta U$ is if the temperature doesn't change, so that they are both equal to zero.