You are perhaps making it a little harder than necessary to take the next step. When you unwrap a recurrence in this way, it’s often a good idea not to combine the terms that don’t involve $T$:
$\begin{align*} T(n)&=2T(n-1)+n\\ &=2\Big(2T(n-2)+(n-1)\Big)+n\\ &=2^2T(n-2)+2(n-1)+n\\ &=2^2\Big(2T(n-3)+(n-2)\Big)+2(n-1)+n\\ &=2^3T(n-3)+2^2(n-2)+2(n-1)+n\;. \end{align*}$
When you leave the intermediate results in this form, it’s very easy to see that after $k$ steps you’ll have
$\begin{align*} T(n)&=2^kT(n-k)+2^{k-1}\big(n-(k-1)\big)+2^{k-2}\big(n-(k-2)\big)+\ldots+2^1(n-1)+2^0n\\ &=2^kT(n-k)+\sum_{i=0}^{k-1}2^i(n-i)\;. \end{align*}$
This makes it easy to see what the final step of the unwrapping will yield, and it also tells you that you’ll need to be able to work out
$\sum_{i=0}^m2^i(n-i)=\sum_{i=0}^m2^in-\sum_{i=0}^m2^ii=n\sum_{i=0}^m2^i-\sum_{i=0}^mi2^i$
for some value of $m$. One of those last two sums is easy, since it’s geometric; the other is a little trickier, but it’s been discussed many times on this site.