The easiest way to see imaginary time used is in elementary quantum mechanics in one dimension. (This is the explanation cribbed from wikipedia).
Suppose we're looking at a tunneling-through-a-barrier problem. We start with the Schrodinger equation:
$$
-\frac{\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2}+V(x)\psi(x) = E\psi(x)
$$
Make the ansatz
$$
\psi(x) = \psi_0 \exp(\frac{i}{\hbar}S(x))
$$
Then we get
$$
-\frac{i\hbar}{2m}\frac{d^2S(x)}{dx^2}+\frac{1}{2m}\left(\frac{dS(x)}{dx}\right)^2+V(x)-E=0
$$
which is nonlinear. We can make progress with an $\hbar$ expansion
$$
S(x)=S_0(x)+\hbar S_1(x)+\frac{\hbar^2}{2}S_2(x)+...
$$
After long calculation we can calculate various amplitudes and derive things like the barrier tunneling coefficient
$$
T=\exp(\frac{2}{\hbar}Im(S))
$$
where
$$
Im(S)=\int_a^b |p(x)|dx
$$
($p(x)=\sqrt{2m(E-V)}$) and $a$ and $b$ are the $x$ values where the potential function is such that $E<V(x)$. Now Feynman offers another way to approach this, namely that the amplitude to get from x=a to x=b is just $$\langle x=b|\exp(\frac{iHt}{\hbar})|x=a \rangle = \int \mathcal{D}[x(t)] \exp(\frac{iS[x(t)]}{\hbar}) \ \ \ (1)$$ where the integral is over the space of classical paths $x(t)$ with the right endpoints. Now this, although very elegant, is extremely hard to compute: it's an integral over an infinite dimesional space after all! The imaginary time trick works as follows:
You just make a change of variable $$t=-i\tau$$ then the action
$$
S(x(t))=\int{\frac{1}{2}}m \left({\frac{dx}{dt}}\right)^2-V(x) dt
$$
becomes $$ S(x(\tau))=i\int{ \frac{1}{2}}m \left({\frac{dx}{d\tau}}\right)^2+V(x) d\tau$$ so the potential energy has swapped sign relative to the kinetic energy (and we picked up an overall i factor). Defining $$ S_E(x(\tau))=\int{(\frac{1}{2}}m ({\frac{dx}{d\tau}})^2+V(x)) d\tau$$, our path integral is now $$\langle x=b|\exp(\frac{-H\tau}{\hbar})|x=a \rangle = \int \mathcal{D}[x(\tau)] \exp(\frac{-S_E[x(\tau)]}{\hbar})\ \ (2)$$ Now the integral will be dominated by classical paths which extremize this action. Whereas an extremal path contributing to (1) would require imaginary energy to tunnel through the potential, which looks like a hill, for (2), the potential hill is now a valley and the corresponding extremal case is just that of a ball rolling down one side of the valley and up the other. Having done your computation in Euclidean space, you then proceed by taking whatever answer you got, and rotating back to Minkowski space.
So much for mechanics. You can do the same trick in field theory, where your path integral is now over classical field configurations. The Euclidean space extremal field configurations are called instantons. Now in your question, Hartle and Hawking were interested in what the equivalent, for the initial conditions of the universe, of "x=a" in our simple example is. Just like in the QM example, they were working in Euclidean time, and wanted their equivalent of "x=b" to be a de Sitter universe. Their guess was that, in the path integral, they should include all Euclidean metrics for spaces with no boundary. Just as our Euclidean extremal paths satisfy the equations of classical mechanics in Euclidean time, so the metrics included in the quantum cosmology path integral would satisfy the classical Euclidean signature Einstein equations.
So to summarize, Euclidean time is a clever trick for getting answers to extremely badly behaved path integral questions. Of course in the Planck epoch, in which the no-boundary path integral is being applied, maybe Euclidean time is the only time that makes any sense. I don't know - I don't think there's any consensus on this.