According to David Tong’s notes on statistical physics on pages 139 to 140, a superheated liquid lies on an isotherm between the spinodial curve and the coexistence curve. Hence, does this mean that the superheated liquid has the same temperature as an otherwise ordinarily boiling liquid in equilibrium with its gas phase?
The superheated liquid has a temperature above an ordinary boiling liquid at the same pressure. That's the "super" part of "superheated".
If you take a cup of water, and boil it thoroughly in a microwave (ten seconds or more), then you'll boil out most (or possibly all) of the dissolved air. This removes nucleation sites for boiling. Now let it calm down, and hit it again for 10-60 seconds, depending on your microwave.
It should be superheated. Dropping a tea bag into it will be exciting (and can lead to scalding -- be careful, and do not hold the cup in your hand when you do this). If it's really superheated, then dropping in the tea bag will make it boil violently. It's hard to describe, but if you look at the surface just right it'll look oddly rippled -- you can tell by looking that it's superheated, but it's definitely a look, not something to be described by words.
Superheated liquid is a liquid which is not undergoing phase transition into gas in bulk (=boiling) even if its temperature is higher (that's why it is called superheated) than the minimum temperature necessary for this phase transition to occur (high enough temperature means high enough vapor pressure which can overcome external pressure). Superheated liquid is metastable state, a strong enough mechanical excitation or introduction of nucleation facilitators (dirt, radiation, charged particles) can destroy it and the phase transition then will occur.