Define $\phi:\mathbb{R}\rightarrow \mathbb{R}$ by $\phi(x)=T(\tau_xRf)$. Then, it is polinomially bounded I want to prove the following proposition:

Let $T\in\mathcal{S}'$ and $f\in\mathcal{S}$, where $\mathcal{S}$ is the Schwartz space.
  Define $\phi:\mathbb{R}\rightarrow \mathbb{R}$ by $\phi(x)=T(\tau_xRf)$, where $Rf$ is the reflexion of $f$ and $\tau_x$ is the translation operator (by $x$).
  Then, $\phi\in C^\infty_{pol}(\mathbb{R})$ and $\forall n\in\mathbb{N},\phi^{(n)}(x) = T^{(n)}(\tau_x Rf)=T(\tau_x Rf^{(n)})$.

The second part of the proposition was not very hard to prove: using the definition the derivative at a point of a function, I proved that $\phi'(x) = T'(\tau_x Rf)=T(\tau_x Rf')$ and by induction the result follows.
Now, my problem is in proving that $\phi$ is polynomially bounded. Let me try and state my confusion as clear as possible:
1 - My teacher defined a polynomially bounded function $P:\mathbb{R}\rightarrow\mathbb{C}$ as 
$$
         \forall\alpha\in\mathbb{N}_0,\exists C(\alpha)>0,\exists N(\alpha): \vert P^{(\alpha)}(x)\vert \le C(\alpha)(1 + x^2)^{N(\alpha)}
$$
but I do not see the motivation behind this, and how it relates with the definition that I found when I browsed online: $f\le P\le g$ where $f,g$ are polynomials.  
2 - The proof that my teacher showed us goes as follows:
     $$ \vert \phi^{(n)} \vert = \vert T^{(n)}(\tau_ xRf) \vert \le M\max_{0\le j, k\le N, s\in\mathbb{R}} \vert s^j(\tau_xRf)^{(k)}(s) \vert = \\ = M\max_{0\le j, k\le N, s\in\mathbb{R}} \vert s^j f^{(k)}(x-s)\vert  =M\max_{0\le j, k\le N} \vert (x-v)^j f^{(k)}(v)\vert $$
(for some $M>0$ and some $N\in\mathbb{N}$, and $v=x-s$). He then wrote "This is a polynomial in $x$". I do not see how, and I do not see how this proves the assertion. Also, I do not know why the first inequality is correct.
 A: The definition your teacher gives is stronger than the one you have found. Let us call the first $(T)$ and the second $(F)$. For example the function $\sin( e^x )$ is polynomially bounded with $(F)$ but not with $(T)$.
However: $f$ satisfies $(T)$ $\iff$ $f^{(n)}$ satisfies $(F)$ for all $n\in\Bbb N$. So while $(F)$ wants your function to be sandwiched by polynomials, $(T)$ wants both the function and all its derivatives to satisfy this property.
The first inequality does not make much sense to me either, indeed I don't see why it is necessary. You can just do the following:
$$|\phi^{(n)}(x)|=|T((-1)^n\tau_x Rf^{(n)})|≤\|T\|\cdot\|\tau_x\|\cdot\|R\|\cdot\|f^{(n)}\|$$
so the expression is actually bounded by a constant. Here we are viewing $\mathcal S$ as a subset of $C_0(\Bbb R)$ for the norms to make sense. (Note that $\|\tau_x\|$ and $\|R\|$ are both $1$ as these are isometries.)
You are also missing a minus sign at some point:
$$\phi'(x)=\lim_{h\to0}\frac{T(\tau_{x+h}Rf)-T(\tau_xRf)}h=\lim_{h\to0}T\left(\tau_x R \frac{\tau_{-h}f-f}h\right)=T(-\tau_xRf')$$
where the limit can be pulled in since all maps are continous and $(\tau_hf-f)/h$ converges in sup norm to $f'$ (this follows from the mean value theorem).
