How do we arrive on this kernel equation? In Feynman and Hibbs, we see the following equation:
$$K(b,a)~=~\sum_{\text{paths from $a$ to $b$}} \phi [x(t)] \tag{2-14}$$
which is valid always.
Now, they write 
$$\phi[x(t)] ~=~ \text{const} ~e^{iS[x(t)]/\hbar} \tag{2-15}$$ 
which is where I have a problem. This "functional" seems to change with situations, for example for a particle moving at the speed of light, and being constrained to move only forward or backward, the "functional" now becomes 
$$\phi ~=~ (i\epsilon)^R.\tag{2-26}$$
So, my question is this. what exactly are these "functionals" that we are talking about, and how do we arrive of the exact form of a functional depending on the situation? When is it in the form of the exponential "functional" as stated above?
 A: The propagator (kernel) can always be expressed as the path integral with a suitably chosen action:
$$
K(b,a) = \int \mathcal{D}[\psi]\ e^{iS[\psi]/\hbar} \ ,
$$
where $\psi$ denotes the configuration of your system.  The subtlety comes in choosing an appropriate $S$ for your situation.
In single-particle, non relativistic quantum mechanics the "configuration" is just the trajectory of your particle $x(t)$ and this becomes equivalent to Eq. 2-15 in Feynman & Hibbs:
$$
K(b,a) = \int \mathcal{D}[x(t)]\ e^{iS[x(t)]/\hbar} \ .
$$
Unfortunately, this exact prescription (describing your particle by its trajectory) doesn't work in relativistic scenarios, because (for instance) your particle can pair produce.  This is mentioned in Section 2-6 of Feynman & Hibbs.  For a relativistic system, you must use quantum field theory.  Your configuration is now defined by a field $\psi(t,\vec{x})$ instead of just a trajectory $x(t)$.  Now the kernel looks like:
$$
K(b,a) = \int \mathcal{D}[\psi(t,\vec{x})]\ e^{iS[\psi(t,\vec{x})]/\hbar} \ .
$$
$S$ is now the action for a relativistic field, for instance $S = \int d^4x \left(\bar{\psi} i\gamma^\mu \partial_\mu \psi - m \bar{\psi}\psi\right)$ for a free fermion.
As pointed out by Qmechanic in the comments, your example $\phi = (i\epsilon)^R$ is the Feynman checkerboard model.  This is a model Feynman came up with to perform path integrals for relativistic particles using only their trajectories, bypassing the field theory.  It was constructed so that the kernel would obey the Dirac equation in the limit $\epsilon \to 0$.  It's a little weird, and only works in 1 dimension (although apparently extensions to multiple dimensions exist). 
In all situations of interest, to the best of my knowledge, kernels/propagators take the form $\sum e^{iS/\hbar} $.  This checkerboard model is the only time I've encountered anything different, and its a pretty unique case.
