My textbook states "A system is in thermodynamic equilibrium if it is in Thermal, Mechanical, and chemical equilibrium simultaneously".
Since you are talking about thermodynamic equilibrium with the surroundings, I would replace that statement (which is fine by itself, though it is not comprehensive) with something like: "A system is in thermodynamic equlilibrium with its surroundings when it is in equilibrium with respect to all forms of work or heat flow that can move across its boundary".
Thus if the boundary is rigid, even a system is in thermodynamic equilibrium, it would not be in mechanical equilibrium with its surroundings, because it is not communicating with its surroundings mechanically. If the boundary is adiabatic, a system at thermodynamic equilibrium would not be in thermal equilibrium with its surroundings, because it is not communicating thermally with its surroundings (e.g., if the walls are adabiatic, the system could be in thermodynamic equilibrium at, say T = 300 K, while the temperature of the surroundings could have any value (yes, above absolute zero, of course).
The Wikipedia article on "reversible processes" states that the system is thermodynamic equilibrium with the surroundings throughout the process. This means, as per my initial statement, $\ce{Tsys=Tsurr}$ at every instant of the process.
Correspondingly, the above is not generally true; it's only true for isothermal reversible processes (you may be misrepresenting the content of the Wikipedia article here by not including the isothermal restriction in your statement—I don't know, since you didn't provide the reference). In an adiabatic reversible process, by contrast, the system is not in thermal equilibrium with the surroundings and, typically, $\ce{Tsys \ne Tsurr}$ (except perhaps at one point, or in the case that you have an adiabatic process with no temperature change, such as the free expansion of an ideal gas).
Since $\ce{Tsys=Tsurr}$ implies that $\ce{Tsurr}$ is not constant as long as the process isn't isothermal.
You have it backwards. For $\ce{\Delta}S_{\text{surr}}=\dfrac{q_{\text{surr}}}{T_{\text{surr}}}$ to apply, the surroundings must be an infinite heat bath. I.e., the surroundings are so vast that their intensive properties (e.g., temperature) are unaffected by exchange of heat, work, or matter with the system. Thus, for isothermal processes in a system that is in thermal equilibrium with its surroundings, it is the temperature of the surroundings that determines the temperature of the system, not the other way around.