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$n$ is free in the expression $\sum_{r=0}^n 2^{n-r}\binom{n+r}{n}$. The only variable not free in that expression is $r$. What $\sum_{r=0}^n$ means is to treat what comes next as a function of $r$, and compute the sum of that function evaluated at $0$, $1$, and so on up to $n$. If $n$ is used inside that function, its value isn't affected by the $\sum$ notation. The fact that it's used in setting one of the limits of the $\sum$ is irrelevant. So if you are dividing the entire expression by some term involving $n$, it is legal to bring that term inside the summation, because
$$
\frac{f(0) + f(1) + \cdots + f(n)}{g(n)} =
\frac{f(0)}{g(n)} + \frac{f(1)}{g(n)} + \cdots + \frac{f(n)}{g(n)}$$

The other point I would make is a subtle one, but it trips people up a lot when learning algebra. In a change of variables, you aren't replacing $r$ with $n - r$. You're introducing a new variable $k = n - r$, and then eliminating $r$ using the rules permitted by algebra.

So:

\begin{align}
\sum_{r=0}^n 2^{n-r}\binom{n+r}{n} &= 2^{2n} \\\\
\sum_{r=0}^n 2^k\binom{n+(n - k)}{n} &= 2^{2n} \quad \text{(simple substitution)} \\\\
\sum_{k=0}^n 2^k\binom{n+(n - k)}{n} &= 2^{2n} \quad \text{($r \in \lbrace 0,\ldots,n\rbrace \Leftrightarrow k \in \lbrace 0,\ldots,n\rbrace$)} \\\\
\sum_{k=0}^n \frac{2^k\binom{n+(n - k)}{n}}{2^{2n}} &= 1 \quad \text{(divide)} \\\\
\sum_{k=0}^n \frac{\binom{2n - k}{n}}{2^{2n - k}} &= 1 \quad \text{(tidy up)}
\end{align}