Thermodynamic process "in thermal equilibrium with the surroundings" 
"During this process, the system is always in thermal equilibrium with the surroundings."

Does this statement mean the process is adiabatic? I would say yes: since the temperature of both systems is the same (because of thermal equilibrium), there will be no heat exchange. Is this correct?
 A: They are talking about a reversible process.  In any process, the temperature of the system always matches the temperature of the surroundings at the interface.  However, in an irreversible process, even though the temperatures at the boundary may match, there are temperature gradients within the system, and the temperature varies with distance from the boundary (i.e., the temperatures within the system are not uniform).  This results in a heat flux at the boundary.  In a reversible process, the process is carried out very slowly, so that the temperature gradient at the boundary is very low.  This means that, to transfer a significant amount of heat, the process has to take place over a much larger time span.  So there is finite heat transfer in the reversible case too, but it occurs a very long time (approaching infinity).  The reversible process is thus not necessarily adiabatic.  Of course, if the boundary is insulated (adiabatic process), the temperature gradient at the boundary is going to be zero, and no heat transfer will occur.
A: Not necessarily, just because two systems have the same temperature does not mean there is no heat flow. This just means there is no net flow of heat.
A: Your question is an excellent one because you have identified what seems to be an inconsistency.
Heat is defined as energy transfer due to temperature difference. As @Chester Miller points out the highlighted statement is talking about a reversible process. A reversible process is an idealization in which the temperature difference approaches zero in the limit so that the process can be reversed returning both the system and its surroundings to their initial states. 
But in order to be heat transfer, there must be a temperature difference. This simply tells us that all REAL processes are irreversible.
So you might ask, what is the usefulness of a so-called reversible process if it can never be realized. The answer is it sets an upper limit to the maximum possible efficiency (that of the Carnot cycle) of any thermodynamic process against which to judge real processes involved in cycles by.
Hope this helps.
A: No! In fact, in order to maintain constant temperature in the system some heat will have to flow between system and surroundings. Here is an example:
Suppose we compress a gas. If we conduct the process adiabatically, the gas will heat up. To keep its temperature constant then, we must cool it, i.e., remove some heat and pass it to the bath. In practice we would have to conduct the expansion slowly to give enough time for the gas temperature to equalize with the bath. 
The misconception that gives rise to your question comes from the intuitive expectation that heat only flows between systems with finite $\Delta T$ between them. It is possible, however, in the limiting sense, to transfer heat between two systems whose temperature difference is $dT>0$, i.e., a vanishing differential, as long as this differential is positive. 
