To understand this question (and its possible answers) properly, we need to discuss a couple of concepts related to **temperature** and its relation to quantum states. Since I think the question makes more sense in the **solid state**, this answer will assume that's what we're talking about.

For starters, I find it useful to think about Boltzmann's distribution: a probability distribution that gives the probability $p_i$ that a system will be in a certain state $i$ as a function of that state’s energy ${\varepsilon}_i$ and the temperature $T$ of the system:

$p_i={\frac{e^{- {\varepsilon}_i / k T}}{\sum_{j=1}^{M}{e^{- {\varepsilon}_j / k T}}}}$

where $k$ is Boltzmann's constant.

In a system that is in **equilibrium**, as defined by statistical mechanics, the population of the different quantum states is governed by this equation (the system will be in a *thermal* state). If we think of a single quantum system rather than a collection or "*ensemble*", this distribution of populations would correspond to a series of weights in a mixed state.
In any case, **these are not the conditions one needs for quantum computing**, where at any given time we want to have a good control the wavefunction. However, note that these probabilities have an exponential dependence on ${\varepsilon}_i$. This will be important further down.

Additionally, we need to consider phonons, the collective excitations in periodic, elastic arrangements of atoms or molecules in condensed matter. These are often the carriers of energy to and from our qubits into the part of the solid where we do not have an exquisite quantum control and thich therefore is thermalized: the so-called *thermal bath*.

Why must quantum computers operate under such extreme temperature conditions?

We can never fully control the quantum state of a solid chunk of matter. At the same time, **we do need full control over the quantum state of our quantum computer**, meaning *the subset of quantum states where our information resides*. These will live in pure states (including quantum superpositions), surrounded by a disordered -thermalized- environment.

Think about the Boltzmann distribution described above, and about the exponential term. In practice, its equation means that we can assume $p_i=0$ when the relation between temperature and the energy of the states we're interested in (which often means the states corresponding to our qubits) is such that ${\varepsilon}_i<<kT$.

Kinetics are often hard to model, but you know that **inevitably your system will tend to thermalize**. So, you need to keep your quantum computer, for as long as possible, in a state such that the only excitations that occur are those corresponding to the quantum states and quantum operations that are part of your computation. If the temperature of the solid where the quantum system is residing is low, you only need to worry about your qubits uncontrollably relaxing to a lower-energy state (which is bad enough). If the temperature is high, you also need to worry about your qubits being uncontrollably excited to higher-energy states. Inevitably, this also includes states that are outside your computational basis, meaning states that, for your qubit state, are neither $|0>$ nor $|1>$, nor any complex combination thereof: harder-to-correct errors.

If you now think about the phonons, recall that they are excitations, which cost energy, and thus are more abundant at high temperature. With rising temperatures, there is a rising number of available phonons, and they will present rising energies, sometimes allowing for interaction with different kinds of excitations (accelerating the kinetics toward thermalization): eventually, those that are detrimental to our quantum computer.

Is the need for extremely low temperatures the same for all quantum computers, or does it vary by architecture?

It does vary, and dramatically so. Within the solid state, it depends on the energies of the states that constitute our qubits. Outside the solid state, as pointed out above and in a follow-up question (Why do optical quantum computers not have to be kept near absolute zero while superconducting quantum computers do?), it's a whole another story.

What happens if they overheat?

See above. In a nutshell: you lose your quantum information faster.