According to Taylor's theorem, one can write $$f(x+h)=\sum_{k=0}^\infty\frac{f^k(x)}{k!}h^k,\tag{1}$$ where $f^k$ is the $k^{th}$ derivative of $f$ at $x$. Let us assume that the series converges for all $x$.
Question 1: Given $D\equiv \frac{d}{dx}$, I suspect $$f(x+D)=\sum_{k=0}^\infty\frac{f^k(x)}{k!}D^k,\tag{2}$$ where $D^{k}$ is applying operator $D$ $k$ times, is a valid operation. Is this correct?
Question 2: Assume that $f(x)=\sum_{0}^\infty a_{k}\psi_{k}(x)$, where $\psi_{k}(x)$ are spline wavelets of order $n$ (i.e., $\frac{d^{n}\psi_{k}(x)}{dx^n}=0$). Is it correct saying that $$f(x+D)=\sum_{0}^\infty a_{k}\psi_{k}(x+D)\tag{3}$$ operator has order $n$? In order words, $f(x+D)x^{n+1}=0$.
Question 3: Given the fact that any $f \in L^2({\mathbb{R}})$ can be represented by the wavelet series above and considering $n=3$ (Quadratic spline), is it correct stating that $f(x+D)x^N$ for $N>2$ is always zeros?
Absolutely not. In general, such a relation only holds for an operator D commuting with x, $[D,x]=0$, which your D is not.
Here is an obvious counterexample, for $f(x)=x^2$, $$ (x+D)^2= D^2 + 2xD+1+ x^2\\ \neq D^2 + 2xD+ x^2, $$ the second line being your Taylor series-inspired false identity.
The other questions are addressed, and indirectly settled in the negative, in your previous question, no? Once you change the domain of your order 3 spline, you are no longer on the real line, nor is its range, and its very definition collapses into ambiguity.