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
$$
\int_1^2 \frac{\delta Q_{\textrm{sys}}}{T_{\textrm{res}}} = \frac{Q_{1\to2}}{T_2}
< Q_{1\to2}\frac{\ln\left({T_2}/{T_1}\right)}{T_2-T_1} = \Delta S_{\textrm{sys},1\to2}.
$$
There are two techniques that I use for simple situations.
First, can I actually imagine how the reverse process would work? Intuitively, it makes sense that I can just push a pendulum back to the original state. But obviously, if I were to put a piece of room temp. metal in room temp. water, they wouldn't suddenly diverge in temperature. This is essentially just playing the system backwards in time, and seeing if it follows your intuition. Knowledge about entropy helps though.
The other technique stems from the mathematicians definition of invertible functions. If multiple initial conditions can result in the same outcome, then the situation isn't invertible. For example, slightly hotter metal and slightly colder water result in the same outcome: so there would be no way to know which original situation to pick if you tried the inverse. A caveat though: it is very hard to tell when situations are exactly identical, as opposed to nearly identical. If they are just nearly identical, then the system could just be chaotic (large dependence on initial conditions) but reversible.
However, these can become difficult to apply in complex scenarios; that's why the entropy definition is so powerful, and why we use it.
Best Answer
You need to consider the surroundings as well. If you go from state A to state B via a reversible process, the change in system's entropy exactly cancels out the opposite change in entropy for the surroundings; so overall there is no change in entropy. On the other hand, if it were an irreversible process, entropy change of the system (though same as the reversible case as it's a state function), doesn't cancel out the entropy change for surroundings. And overall there is a positive change in universe's entropy.
Further, in the reversible case, you can directly relate the change in system's entropy with heat transferred reversibly divided by temperature of the surrounding. But in irreversible case, the heat transferred irreversibly can't be used to evaluate entropy change and you would need to use an equivalent reversible process with same equilibrium end points.