Exactly as you said: "(5) seems to say that the entropy change for an irreversible process is 0 for the case of the system itself, but is > 0 for the case of the surroundings." After the cycle is finished, by definition of the "cycle", the system returns to its original state irrespective of all the irreversibilities that may have taken place within it during some portions of the cycle.
Entropy being a state variable it also assumes its initial value after the cycle. The only place therefore where the entropy will have increased after the cycle is in the reservoirs that do not perform any "cycle", they are sinks and sources of heat and their state can and will change per Clausius., i.e., $$\Delta S_{res} =\frac{q_{res}}{T_{res}}$$
As regards to equality vs. inequality I always learned and thought I understood that any time there is irreversible process it is inequality and never equality, and that goes for any part of the process. This can be understood intuitively by noting that irreversibility is caused by internal dissipation, i.e., heat generation, therefore during irreversible processes less heat is absorbed from the reservoir and more heat is rejected to the reservoir if compared to a reversible transition.
Incidentally, the absorbed or rejected $q$ of a reservoir is always reversible, that is why its entropy change is just the ratio of the heat and its temperature.
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
This is basically the application of the Clausius inequality to an irreversible cycle. The zero on the right hand side of the equation represents the change in entropy of the working fluid over the cycle, which is equal to zero (since the initial and final states around a complete cycle are the same). On the left-hand side, the $T_i$ represents the temperature at the interface between the working fluid and its surroundings, at which the heat transfer $Q_i$ is occurring. The proper statement of the Clausius inequality always requires you to use the temperature at the interface where the heat transfer is occurring.
ADDENDUM During an irreversible cycle, all the entropy generated within the system (by irreversibility) in each cycle $\delta$ is transferred from the system to the surroundings, so that the change in entropy of the system in each cycle is zero. The entropy transferred from the surroundings to the system during a cycle is given by $\sum_{i=1}^n\frac{Q_{i}}{T_{i}}$, so the entropy transferred from the system to the surroundings during the cycle is $\left(-\sum_{i=1}^n\frac{Q_{i}}{T_{I}}\right)$ That means that $$\delta =\left(-\sum_{i=1}^n\frac{Q_{i}}{T_{I}}\right)$$or, equivalently, $$\sum_{i=1}^n\frac{Q_{i}}{T_{I}}=-\delta$$Since the irreversible generation of entropy is always positive, this means that the left hand side of this equation for a cycle is negative, or $$\sum_{i=1}^n\frac{Q_{i}}{T_{I}}\leq0$$