Clarification on usage of dU =nCvdT
The reason why statement #2 doesn't make sense to you is because it is not generally true!
For a closed system of constant composition, with P-V work only,
$$dU = \left(\frac{\partial U}{\partial T}\right)_V dT +  \left(\frac{\partial U}{\partial V}\right)_T dV = C_v dT + \left(\frac{\partial U}{\partial V}\right)_T dV$$
The difference between that statement and your statement is the extra term at the end, which is the amount by which U changes with V at constant T, times the change in V.
[Minor note: I've dropped the &quot;n&quot; you use, as I'm using total heat capacities, not molar heat capacities.]
However, if we have an ideal gas, U is independent of V, which means:
$$\left(\frac{\partial U}{\partial V}\right)_T=0,$$
Hence, with an ideal gas, the term on the right goes away, giving us your statement #2:
$$dU = C_v dT$$
I.e., your statement #2 doesn't follow directly from #1.  Instead, it's specific to ideal gases.
So:  If a process invoving an ideal gas is carried out at constant pressure instead of constant volume, it is true that $dV$ may not be zero.  But since that non-zero $dV$ doesn't matter when it comes to calculating $dU$ for ideal gases, we can use $C_V dT$ even when $dV \neq 0$.