Answering your questions
(1) As long as the irreversibility arises solely as a consequence of the fact that the system and the environment are at different temperatures, the methods outlined below work. You can calculate $\Delta S_{\textrm{sys}}$ as normal, and calculating $\Delta S_{\textrm{env}}$ is also straight-forward if we treat its temperature as constant. There is no such thing as $\delta Q^{\textrm{rev}}$ and $\delta Q^{\textrm{irr}}$. The difference between the reversible and irreversible cases is the path that the environment takes through state space.
(2) This depends on what $\delta Q$ and what $T$ you're talking about. If $Q$ is the heat flow into the system, and $T$ is the temperature of the system, then this is $\mathrm dS_{\textrm{sys}}=\delta Q_{\textrm{sys}}/T_{\textrm{sys}}$. If these are the heat flow into the environment and the temperature of the reservoir, then this is $\mathrm dS_{\textrm{env}}=\delta Q_{\textrm{env}}/T_{\textrm{env}}$. If $Q$ is the heat flow into the system and $T$ is the temperature of the environment, then $\delta Q_{\textrm{sys}}/T_{\textrm{env}}$ is just $-\mathrm dS_{\textrm{env}}$, and we can interpret the quantity $\mathrm d\sigma = \mathrm dS_{\textrm{sys}} - \delta Q_{\textrm{sys}}/T_{\textrm{env}}$ as the entropy production of this part of the process. In the case where the irreversibility arises solely as a consequence of heat flow between system and environment when they have different temperatures, and if the system operates on a quasi-static cycle, then the net entropy production $\sigma = \oint\mathrm d\sigma$ goes into the environment.
(3) Let's write the Clausius' inequality carefully as
$$
\frac{\delta Q_{\textrm{sys}}}{T_{\textrm{env}}} < \mathrm dS_{\textrm{sys}}.
$$
Using the answer to part (2) and this form of the inequality, I think that dissolves question (3), but I'm not sure.
Now, I think it's worth expanding on these comments:
Preliminaries
(A) If we are talking about the system following the cycle shown above, there is no such thing as $\delta Q^{\textrm{rev}}$ vs $\delta Q^{\textrm{irr}}$. The reason is that in order to even draw that diagram to begin with, we are assuming that the system is undergoing quasi-static processes. The irreversibility is solely a product of the energy exchange with the environment. In particular, it is due to heat flow between the system and its environment when the there is a finite temperature difference between them.
(B) The Clausius' inequality is subtle. The temperature that shows up in $\delta Q/T$ is the temperature of the boundary of the system, not the system itself! In other words, $T$ appearing in the Clausius' inequality is actually the temperature of the environment. This is why during an irreversible process, the entropy change of the system, defined by $\oint \delta Q_{\textrm{sys}}/T_\textrm{sys}$, can be zero, while $\oint \delta Q_{\textrm{sys}}/T_\textrm{res}<0$.
In any case, it is useful to do some calculations explicitly. Let's concentrate on the isochoric process $1\to2$ for the purposes of illustration.
Heuristics
Below, we carefully compute the entropy changes for both system and environment, but for now, let's give a quick heuristic explanation of what's going on.
If---as illustrated in the figures above---the system undergoes a quasi-static process (meaning that the system moves through a sequence of equilibrium states and so always has a well-defined set of thermodynamic variables), then the entropy change of the system is given by integrating $\delta Q_{\textrm{sys}}/T_\textrm{sys}$ from point 1 to 2 along a reversible path, regardless of whether the actual process is reversible or not. If the process is not quasi-static for the system, it is possible that the system can be broken up into subsystems that do undergo quasi-static processes.
In general, one can calculate the entropy change during an irreversible process between two equilibrium states by imagining a quasi-static process between them and calculating $\Delta S$ for that process. If the process is quasi-static, we can use $dS = \delta Q/T$. If not, we can use the thermodynamic relation
$$\mathrm dU = T\,\mathrm dS-p\, \mathrm dV+\mu\, \mathrm d N$$
by solving for $\mathrm \,dS$ and integrating along the reversible path.
Here, we assume that the irreversibility arises solely as a consequence of heat exchange between the system and its environment while they are different temperatures, which means that the system and environment each undergo separate quasi-static processes, but we can think of them as two subsystems comprising a closed system that does not undergo a quasi-static process.
We do a sample calculation carefully below, but note that $T_\textrm{sys}$ is changing throughout the process. On the other hand, the entropy change of the environment is given by integrating $\delta Q_{\textrm{env}}/T_\textrm{env}$ along a reversible path, where these are now quantities associated with the environment.
Now, consider the case where the system is in contact with a single reservoir of temperature $T_2$ throughout this process, which means that at all times, $T_\textrm{env} > T_\textrm{sys}$. In any small part of the process, the heat flow out of the reservoir is equal to the heat flow into the system, and so the entropy gain of the system is necessarily larger than the entropy loss of the reservoir:
$$
\mathrm dS_{\textrm{sys}} = \frac{\delta Q_{\textrm{sys}}}{T_{\textrm{sys}}} > \frac{\delta Q_{\textrm{sys}}}{T_{\textrm{res}}} = -\frac{\delta Q_{\textrm{res}}}{T_{\textrm{res}}} = -\mathrm dS_{\textrm{res}}
$$
Finally, if we were to calculate part of the Clausius' inequality integral, it would be exactly
$$
\frac{\delta Q_{\textrm{sys}}}{T_{\textrm{res}}} = -\mathrm dS_{\textrm{res}} < \mathrm dS_{\textrm{sys}}
$$
as it's supposed to.
Careful calculation
The entropy change of the system is given by
$$
\Delta S_{\textrm{sys},1\to2} = \int_{1}^{2}\frac{\delta Q_{\textrm{sys}}}{T}
= \int_{T_1}^{T_2}\frac{nC_V\,\mathrm dT}{T},
$$
where $C_V$ is the molar specific heat of the gas at constant volume. This evaluates to
$$
\Delta S_{\textrm{sys},1\to2} = nC_V\ln\left(\frac{T_2}{T_1}\right),
$$
which can be written as
$$
\Delta S_{\textrm{sys},1\to2} = Q_{1\to2}\frac{\ln\left({T_2}/{T_1}\right)}{T_2-T_1},
$$
where $Q_{1\to2}$ is the heat flow into the system during this process; this quantity is positive since $T_2 > T_1$.
Now, suppose that this process comes about due to the system being in contact with a thermal reservoir of constant temperature $T_2$. Then, the change in entropy of the reservoir is given by
$$
\Delta S_{\textrm{res},1\to2} = \int_{1}^{2}\frac{\delta Q_{\textrm{res}}}{T_{\textrm{res}}}
= \int_{1}^{2}\frac{-\delta Q_{\textrm{sys}}}{T_2},
$$
assuming that the system and reservoir are otherwise isolated from the rest of the universe so that $\delta Q_{\textrm{res}} = -\delta Q_{\textrm{sys}}$. This last term evaluates to
$$
\Delta S_{\textrm{res},1\to2} = -\frac{Q_{1\to2}}{T_2},
$$
and so the total entropy change of the universe is
$$
dS = \Delta S_{\textrm{sys},1\to2} + \Delta S_{\textrm{res},1\to2}
=Q_{1\to2}\left(\frac{\ln\left({T_2}/{T_1}\right)}{T_2-T_1}-\frac{1}{T_2}\right).
$$
It is relatively straight-forward to show that this quantity is positive for $T_2>T_1$ (our assumption).
The piece of the Clausius' inequality here is then just
$$
\int_1^2 \frac{\delta Q_{\textrm{sys}}}{T_{\textrm{res}}} = \frac{Q_{1\to2}}{T_2}
< Q_{1\to2}\frac{\ln\left({T_2}/{T_1}\right)}{T_2-T_1} = \Delta S_{\textrm{sys},1\to2}.
$$
Best Answer
You went wrong in interpreting the Kelvin-Plank statement of the second law. There is no violation. The statement is (the details of the statement may differ depending on the references)(italics done by me for emphasis):
"It is impossible to devise a cyclically operating heat engine, the effect of which is to absorb energy in the form of heat from a single thermal reservoir and to deliver an equivalent amount of work"
No work is done by the system in the free expansion. In this case no heat is absorbed by the system in the cycle, only rejected by the system to the surroundings as a result of the isothermal compression done by the surroundings needed to return the system to its original state. The system does not deliver work. The surroundings does work on the system. There is no violation.
You can look at this also from the perspective of the surroundings. Since the system is isolated from the surroundings during the free expansion of the system, the surroundings does not partake in that process. When the surroundings does work to compress the gas in the system, it performs work, But this work is not part of a cycle. It is simply a single process, namely, a reversible isothermal process. Net work done during a single process using heat from a single reservoir (as opposed to a sequence of processes comprising a cycle) does not violate the law.
But the whole universe has not been brought back to its original state.
The system has been brought back to its original state (including original entropy) but the state of the surroundings has now been changed. Heat has been transferred to the surroundings increasing its entropy. Thus there is a positive total entropy change of the universe (system + surroundings) making the entire process irreversible.
The following is in response to your follow up questions:
If by the term another "variable" you mean another property, the short answer is there is no other property that I am aware of to account for a process being irreversible than the property of entropy. This is the reason the second law and its associated property, entropy, was needed. The first law (conservation of energy) is satisfied by a process even if the process is impossible, as long as energy is conserved. The simplest example is that of natural heat flow.
We know that heat only flows naturally, or spontaneously (without external influence), from a hot body to a cold body. The reverse has never been observed to occur spontaneously, even though the reverse process does not violate the first law of thermodynamics if the heat lost by the cold body equals the heat gained by the hot body.
The example of the free expansion is more subtle. But the idea of spontaneity applies. The gas spontaneously expands from its chamber to the evacuated chamber. We would never see all the gas that expanded into the evacuated chamber spontaneously return to its original chamber. To be more precise, the probability is essentially zero. But there is no violation of the first law.
Good question. But if the temperature of the environment increases, and only heat transfer to the environment is involved, that would mean the internal energy of the environment would also increase. But that would be impossible in this example.
The work done by the environment on the system to return it to its original state exactly equals the heat transferred from the system to the environment (being that compression of the gas is isothermal). That means the change in internal energy of the environment has to be zero. For the entire cycle the change in internal energy of both the system and the environment is zero. This can be so regardless of whether the cycle is reversible or irreversible. The property of entropy is needed to show irreversibility.
Hope this helps.