I can answer your last question, at least. The derivative acts as a shift operator on Taylor series, so the operator $\frac{d}{dx} - 1$ acts as the forward difference on Taylor series. So their eigenvectors are basically the same; the eigenvectors of the derivative are the exponential functions $e^{\lambda x}$, which have Taylor coefficients $a_n = \lambda^n$, and these are the eigenvectors of the forward difference. I'll think about your other questions.
Edit: Here's a possible way to get a notion of "fractional forward difference." If we write the forward difference operator $\Delta f(n) = f(n+1) - f(n)$ as $D - 1$ where $D$ is the shift operator, it follows that
$\displaystyle \Delta^k f(n) = (D - 1)^k f(n) = \sum_{i=0}^{k} {k \choose i} (-1)^{k-i} f(n+i)$
by the binomial theorem. So a possible extension to non-integer values of $k$ is to use the generalized binomial theorem formally in the above expansion. Unfortunately, the above sum is then infinite and therefore not guaranteed to converge. I'm not really sure how useful or interesting this is.
Edit 2: Well, that doesn't work for a stupid reason; $\sqrt{D - 1}$ doesn't have a Taylor series expansion at $0$. However, the backward difference operator $\nabla f(n) = f(n) - f(n-1)$ can be written as $1 - E$ where $E$ is the other shift operator and $(1 - E)^k$ has a Taylor series expansion for every $k \in \mathbb{C}$. (Note in particular the case $k = -1$.)