Background
The form $d...$ is a differential. It means an infinitesimal change in something as it undergoes a process. The something of interest here is a thermodynamic state function of a system, its surroundings, or the universe. The change in the universe is the sum of the changes in the system and its surroundings, so only two of the three are independent.
Any change in any thermodynamic state function is always independent of the path taken. Two types of paths are defined: reversible and irreversible. The former is defined as a path where the system and its surroundings are in exact thermodynamic equilibrium at all points during the process. The latter is anything else.
Entropy
One definition of entropy change is $dS = \delta q_{rev}/T$, where $\delta q$ is the heat flow that occurs during the process and $T$ is the temperature of either the system OR the surroundings at the instant of the infinitesimal change. We include the subscript $_{rev}$ on heat flow to acknowledge that, while heat flow is path dependent, entropy change is not. We choose a reversible path because it is typically easier to evaluate than an irreversible path. Writing the expression without the $_{rev}$ subscript on heat flow is common but ambiguous.
For a reversible process, the entropy change of the system and the surroundings are equal and opposite at any step. We need only calculate one or the other. The sum is zero. Hence the statement that a reversible process results in $\Delta S_{universe} = 0$ always. This means, we only need to know one or the other, not both.
We can and will always evaluate the (infinitesimal) entropy change of a system $dS_{sys}$ by choosing a reversible process. How then do we determine entropy change during an irreversible process?
The immediate response back is to clarify that we must also define the exact status of the universe when we propose that a system will undergo an irreversible process. Why? Because the exact nature of the path that is irreversible depends on the exact nature of the differences between the system and the surroundings. By simple analogy, the closer the system and the surroundings are to being at exact thermodynamic equilibrium during the proposed process, the smaller will be the irreversibility of the process that occurs.
The magnitude of the irreversible entropy change for any process depends on the magnitude of the difference between the system and the surroundings from exact thermodynamic equilibrium.
We cannot say that the value (magnitude) of $dS_{sys}$ indicates irreversibility by itself. Its value is always evaluated along a reversible path because entropy is a state function. We also cannot say that the value (magnitude) of $dS_{surr}$ indicates irreversibility by itself. Otherwise, we could just invert our definitions of system and surroundings and find ourselves in a paradox. We can only determine whether a process is or is not (was or was not) irreversible when we consider $dS_{univ}$. In that case, we find $dS_{univ} > 0$ indicates a (spontaneous) irreversible process is occurring (has occurred) along the path.
So, if we always determine $dS_{sys} = \delta q_{rev,sys}/T_{sys}$, how do we make sense of the statement that "... $dS > \delta q/T$ indicates an irreversible process?"
We do so by recognizing that this type of statement is a gross simplification. The true statement is this combination:
A reversible process is one where $dS_{univ} = 0$, while $dS_{univ} > 0$ indicates a (spontaneous) irreversible process.
In a reversible process, the entropy change of the system and surroundings are equal and opposite. In an irreversible process, we generate extra entropy. We can assign that "extra" irreversible entropy either to the system or to the surroundings. In truth, the irreversible entropy generated belongs to the process, not to a system or the surroundings.
So, where does that irreversible entropy appear at the molecular level? It appears in the fact that the system and the surroundings are at distinctly different statistical configuration levels during the process. By example, imagine a classic example of an ideal gas that fills half of an insulated chamber while the other half is a vacuum. We break an invisible wall between the two half chambers. The resulting process is irreversible. The entropy change of the system can be calculated along a reversible path. The entropy change of the surroundings is zero because no heat flows into or out of the surroundings during the process. The net change of the universe (system + surroundings) is positive. At the molecular level, we anticipated that breaking the wall would result in an irreversible process because the initial system is an ideal gas going into a "surroundings" of a vacuum. The two locations (ideal gas and vacuum) have distinctly different configurational entropy values.
Gradients in the configurational entropy states within a system can be a source of irreversible processes.
Further discussion of entropy is found at these related postings on this forum.
Trying to understand entropy as a novice in thermodynamics
Variation of entropy with volume under constant pressure
Entropy generation of constant volume heat addition process
Gibb's Energy
As it ends with "... makes no sense to me", your last question shows a need to read further in a textbook. These links may also offer insights.
Does positive total entropy generation means that the entropy generation of the system and the entropy generation of the surroundings are positive too
Why is choosing a suitable thermodynamic potential important?
Best Answer
Let me start by saying I agree with what both @Barbaud Julien and @Chester Miller said. So my answer is only intended to provide a different perspective.
First, it is true that the entropy is a state function of a system, and therefore the change of entropy of the system between any two given equilibrium states will be the same whether or not the process between the states is reversible. However, the change in entropy of the system plus surroundings will depend on the process. It will be zero if the process is reversible, and greater than zero if irreversible.
Consider a cycle consisting of two isothermal and two adiabatic processes. If the processes were all reversible, we would obviously have a Carnot cycle. Since your question focuses on the effect of heat transfer on entropy, let’s assume the adiabatic processes are reversible adiabatic (isentropic) and consider only the impact of the heat transfer processes being reversible or irreversible.
Let’s take the example of a heat addition expansion process of an ideal gas between two equilibrium states. Let the initial and final temperatures be equal. Consequently, from the ideal gas law, the initial and final pressure-volume products are equal. Clearly the process between the states may be a reversible isothermal process. However, it also need not be. We’ll consider both.
Let the temperature of the system be $T_{sys}$ and the temperature of the surroundings be $T_{sys}+dT$. Consider the system and surroundings as thermal reservoirs, i.e., heat transfer between them does not alter their temperatures so that the heat transfer occurs isothermally.
Let a specific quantity of heat, $Q$, transfer from the surroundings to the system. The resulting entropy changes are:
$$\Delta S_{sys}=\frac{+Q}{T_{sys}}$$
$$\Delta S_{surr}=\frac{-Q}{T_{sys}+dT}$$
The net change in entropy (system + surroundings) is thus:
$$\Delta S_{net}=\frac{+Q}{T_{sys}}+\frac{-Q}{T_{sys}+dT}$$
Now, note that if $dT\to 0$, then $\Delta S_{net}\to 0$ and the process is a reversible isothermal process.
However, for any finite temperature difference, $dT>0$, $\Delta S_{net}>0$ and the process is irreversible.
For both the reversible and irreversible processes, the change in entropy of the system is the same. However, for the irreversible process the change in entropy of the system is greater than the change in entropy of the surroundings. The excess entropy that is created in the irreversible expansion process means more heat must be rejected to the cold temperature reservoir (surroundings) during the isothermal compression in order for the cycle entropy to be zero. That results in less energy to do work in the cycle.
Although in this example the same amount of heat is transferred reversibly and irreversibly, clearly the rate of heat transfer will be greater for the irreversible than the reversible process owing to the finite temperature differential for the irreversible process. So for the amount of heat transfer to be the same, the product of the very slow heat transfer rate times the very long time for the reversible process would need to equal the product of the higher heat transfer rate times the shorter time duration for the irreversible process.
Hope this helps.