The ideal gas equation only applies to an ideal gas in a thermodynamic equilibrium state, and not to one that is rapidly deforming. The 2nd equation you wrote applies only to two closely neighboring thermodynamic equilibrium states of a system, and describes the relationship between the differential changes in U, S, and V between these adjacent states. Neither equation describes the intermediate changes along an irreversible path between two widely separated thermodynamic equilibrium states. So how do you determine the change in entropy between these two end states that are connected in practice by an irreversible path?
Step 1: This is the most important step. Completely forget about the actual irreversible path between the two end states. Focus only on the initial and final end states.
Step 2: Devise a reversible path between the two end states. The reversible path you devise does not have to bear any resemblance whatsoever to the actual irreversible path, as long as it starts and ends at the same two end states. A reversible path consists of a continuous sequence of thermodynamic equilibrium states. So, along the path you choose, the system can be no more than slightly removed from thermodynamic equilibrium at every point along the path. There are an infinite number of reversible paths than can take you from state A to state B. So choose one that is convenient for performing step 3.
Step 3: Calculate the integral of dQ/T for the reversible path that you devised in step 2.
When they say that the change in entropy is equal to the integral of $dQ_{rev}/T$, what they mean is that you have to calculate the integral for a reversible path.
You can alternatively use your equation 2 to determine the change in entropy since it automatically implies a reversible path.
Suppose we consider the Van der Waals gas as more accurate model for a real gas. It can be shown that its internal energy is
$$
U(T,V, N) = cNT - \frac{aN^2}{V}, \qquad c>0,\quad a>0.
$$
Since for an adiabatic free expansion $Q = W = 0$, we have $\Delta U = 0$.
It follows that for a gas of fixed particle number
\begin{align}
0 = U_f - U_i = cN(T_f-T_i) - aN^2\left(\frac{1}{V_f} -
\frac{1}{V_i}\right),
\end{align}
and therefore
$$
\Delta T = \frac{a}{c}N\left(\frac{1}{V_f} -
\frac{1}{V_i}\right).
$$
There is a nonzero change in temperature that depends on the increase in volume. This expression can be made more immediately informative if we define $x = V_f/V_i$ so that $V_f = xV_i$ and
$$
\Delta T = \frac{a}{c}\frac{N}{V_i}\left(\frac{1}{x} - 1\right).
$$
If the gas expands then $x>0$, and the temperature decreases.
Edit. More General Considerations
Assuming the virial expansion (which amounts to quite a general equation of state, we have
$$
\frac{PV}{NkT} = 1 + B(T)\frac{N}{V} + C(T)\left(\frac{N}{V}\right)^2 + \cdots
$$
It follows that
$$
\left(\frac{\partial U}{\partial V}\right)_{T,N}
= T^2\left(\frac{\partial(P/T)}{\partial T}\right)_{T,N}
= kT^2\frac{N}{V}\left(B'(T)\frac{N}{V} + C'(T)\left(\frac{N}{V}\right)^2 + \cdots\right)
$$
For gases that are not too dense, we can truncate at the second virial coefficient $B(T)$ term to good approximation, so the sign of this derivative is the same as the sign of $B'(T)$. For many real gases, the second virial coefficient is a monotonically increasing function of temperature (at least for temperatures that are not too high, see diagram below), so $B'(T) > 0$ and thus to good approximation
$$
\left(\frac{\partial U}{\partial V}\right)_{T,N} \gtrapprox 0.
$$
Best Answer
The reason that the temperature changes in an expansion/compression of a real gas at constant internal energy can be quite intuitively understood.
Suppose that the interactions between the gas molecules are repulsive. If we compress the gas while fixing its energy, the average intermolecular distance decreases, leading to the increase in the potential energy. Accordingly, the kinetic energy, as well as the temperature, has to decrease (energy conservation).
Similar lines of reasoning apply for other cases, i.e., expansion or compression of a gas with repulsive or attractive interactions.