In that last line you really want
$$\frac{2}{n}\left(\left(-2+\frac{2k}{n}\right)^2 + \left(-2+\frac{2k}{n}\right)\right)\;,\tag{1}$$
since the factor of $\dfrac2n$ applies to both terms. In fact, you really want a sum,
$$\sum_{k=1}^n\frac{2}{n}\left(\left(-2+\frac{2k}{n}\right)^2 + \left(-2+\frac{2k}{n}\right)\right)\;,$$
but let’s simplify $(1)$ a bit first.
$$\begin{align*} \left(-2+\frac{2k}{n}\right)^2 + \left(-2+\frac{2k}{n}\right)&=\left(4-\frac{8k}n+\frac{4k^2}{n^2}\right)+\left(-2+\frac{2k}n\right)\\ &=2-\frac{6k}n+\frac{4k^2}{n^2}\;, \end{align*}$$
so $(1)$ is
$$\frac2n\left(2-\frac{6k}n+\frac{4k^2}{n^2}\right)\;,$$ and we want to evaluate
$$\sum_{k=1}^n\frac2n\left(2-\frac{6k}n+\frac{4k^2}{n^2}\right)\;.$$
Now $\dfrac2n$ doesn’t depend on $k$, so you can factor it out of the summation to get $$R_n=\frac2n\sum_{k=1}^n\left(2-\frac{6k}n+\frac{4k^2}{n^2}\right)\;.\tag{2}$$ The trick now is to break up the summation into three simpler ones,
$$\sum_{k=1}^n\left(2-\frac{6k}n+\frac{4k^2}{n^2}\right)=\sum_{k=1}^n2-\sum_{k=1}^n\frac{6k}n+\sum_{k=1}^n\frac{4k^k}{n^2}\;,\tag{3}$$
and evaluate each of those separately. The first one is just $2+2+\ldots+2$, with $n$ terms, so it’s $2n$. The other two you can simplify as you’ve done before to $$\frac6n\sum_{k=1}^nk\quad\text{and}\quad\frac4{n^2}\sum_{k=1}^nk^2\;.$$ Now you want to evaluate these, substitute back into the righthand side of $(3)$, and plug that back into $(2)$ to get $R_n$. Then see if you can calculate $\lim\limits_{n\to\infty}R_n$.
Added: First,
$$\frac6n\sum_{k=1}^nk=\frac6n\cdot\frac{n(n+1)}2=3(n+1)\;,$$
and
$$\frac4{n^2}\sum_{k=1}^nk^2=\frac4{n^2}\cdot\frac{n(n+1)(2n+1)}6=\frac{2(n+1)(2n+1)}{3n}\;,$$
so $(3)$ is $$2n-3(n+1)+\frac{2(n+1)(2n+1)}{3n}=-n-3+\frac{4n^2+6n+2}{3n}\;,$$ and $$\begin{align*}R_n&=\frac2n\left(-n-3+\frac{4n^2+6n+2}{3n}\right)\\ &=-2-\frac6n+\frac{8n^2+12n+4}{3n^2}\\ &=-2-\frac6n+\frac83+\frac4n+\frac4{3n}\;. \end{align*}$$
Now just take the limit as $n\to\infty$.