I want to show:
We can use the two equivalent definitions of rapidly decreasing functions: $x^nf(x)$ is bounded for all $n\geq0$ or The limit as x approaches positive or negative infinity of $x^nf(x)$ is $0$ for all $n\geq0$ . Using the first definition, it is equivalent to saying that $|(x-y)^nf(x-y)|\leq$M is bounded. So we can just expand using the binomial theorem. Now I am stuck at this point because I am not sure how the term $(1+y)^k$ even appears.
The general case
In general, it is useful to use the weight $\langle x\rangle = \sqrt{1+x^2}$. Then it is not difficult to show that $\langle x+y\rangle \leq \sqrt 2\, \langle x\rangle\langle y\rangle$ (taking the square on both sides) which is called Petree's inequality, and to deduce that $\langle x-y\rangle^{-1} \leq \sqrt 2\, \langle x\rangle^{-1} \langle y\rangle$ (by changing $x$ by $x-y$). Therefore, since $\langle x\rangle^n \,|f(x)| \leq (1+|x|)^n \,|f(x)| \leq 2^{n-1} (M_0+M_n)$ you deduce that $$ |f(x-y)| \leq C_n\,\langle x\rangle^{-n}\, \langle y\rangle^n \leq \frac{C_n}{|x|^n} \,(1+|y|)^n $$ with $C_n = 2^{3n/2-1} (M_0+M_n)$.
Using Binomial theorem when $n$ is an integer
If $n$ is an integer, you can indeed use the binomial theorem. Notice that $$ \left|\sum_{k=0}^n \binom{n}{k} x^{n-k}\,(-y)^{k} \,f(x-y)\right| \leq M_n $$ implies by the triangle inequality $$ \left|x\right|^n \left|f(x-y)\right| \leq M_n + \sum_{k=1}^n \binom{n}{k} \left|x^{n-k}\,y^{k} \,f(x-y)\right| $$ Now, for any $k\in\{1,\dots,n\}$ and any $\varepsilon_k>0$, by Young's inequality for the product $a^\theta\, b^{1-\theta}\leq \theta\, a+(1-\theta)\,b$ with $a=\varepsilon_k \,|x|^n$ and $b=\varepsilon_k \,|y|^n$, it holds $$ \left|x^{n-k}\,y^{k}\right| = \left|(\varepsilon_k^{1/n}\, x)^{n-k}\,(\varepsilon_k^{-(n-k)/(kn)}\,y)^{k}\right| \leq \varepsilon_k\,\frac{n-k}{n} \left|x\right|^n + \varepsilon_k^{1-n/k}\,\frac{k}{n} \left|y\right|^n $$ and so from the above formula, and since $f$ is bounded (say $|f|\leq M_0$) then $$ \left|x\right|^n \left|f(x-y)\right| \leq M_n + \left(\sum_{k=1}^n \binom{n-1}{k}\,\frac{n-k}{n}\, \varepsilon_k\right) \left|x\right|^n \left|f(x-y)\right| + \left(\sum_{k=1}^n \binom{n}{k}\,\,\frac{k}{n}\, \varepsilon_k^{1-n/k}\right) \left|y\right|^n\, M_0 \\ \leq M_n + \left(\sum_{k=1}^{n-1} \binom{n-1}{k} \varepsilon_k\right) \left|x\right|^n \left|f(x-y)\right| + \left(\sum_{k=0}^{n-1} \binom{n-1}{k} \varepsilon_k^{1-n/k}\right) \left|y\right|^n\, M_0 $$ and so it suffices to take $\varepsilon_k$ sufficiently small so that the first sum is smaller than $1$. For example if you take $\varepsilon_k$ of the form $\varepsilon_k = (c-1)^k$ for a fixed (possibly $n$ dependent) constant $c > 1$ then you can compute the sums using the binomial formula and you get $$ \left(2-c^{n-1}\right)\left|x\right|^n \left|f(x-y)\right| \leq M_n + \frac{c^{n-1}}{(c-1)^n} \left|y\right|^n\, M_0 $$ and so if $c\in (1,2^{1/(n-1)})$, then $2-c^{n-1}>0$ and so $$ \left|f(x-y)\right| \leq \frac{1}{\left|x\right|^n} \left(\frac{M_n}{\left(2-c^{n-1}\right)} + \frac{c^{n-1} M_0}{(c-1)^n\left(2-c^{n-1}\right)} \left|y\right|^n\, \right) $$ For example, taking $c = 2^\frac{1}{2(n-1)}$ leads to $$ \left|f(x-y)\right| \leq \frac{1}{\left|x\right|^n} \left(\frac{M_n}{\left(2-\sqrt 2\right)} + \frac{\sqrt 2\, M_0}{(c-1)^n\left(2-\sqrt 2\right)} \left|y\right|^n\, \right) < \frac{C_n}{|x|^n} \left(1+|y|\right)^n $$ with $C_n = \left(2\,M_n+3\,M_0\right)\left((\sqrt 2)^\frac{1}{n-1}-1\right)^{-n}$.
Remark: the constant seems however much worse as $C_n \sim \left(2\,M_n+3\,M_0\right) \left(\tfrac{n}{\sqrt 2-1}\right)^n$.