Proving $4^n=\sum_{k=0}^n2^k\binom{2n-k}{n}$

A solution through Complex Analysis and the residue theorem.

$$\mathcal{S}(n)=\sum_{k=0}^{n} 2^k\binom{2n-k}{n} = \sum_{k=0}^{n}2^{n-k}\binom{n+k}{n} $$ is the coefficient of $x^n$ in the product between $\sum_{k\geq 0}2^k x^k=\frac{1}{1-2x}$ (geometric series) and $\sum_{k\geq 0}\binom{n+k}{n}x^k = \frac{1}{(1-x)^{n+1}}$ (stars and bars). In particular

$$ \mathcal{S}(n)=\text{Res}\left(\frac{1}{(1-2x)\left[x(1-x)\right]^{n+1}},x=0\right) $$ but due to the symmetry of the meromorphic function $\frac{1}{(1-2x)\left[x(1-x)\right]^{n+1}}$ the residue at $0$ and the residue at $1$ are the same number. The only other pole is at $x=\frac{1}{2}$ and the sum of the residues is zero, hence $\mathcal{S}(n)=4^n$ can be proved from the straightforward $$ \text{Res}\left(\frac{1}{(1-2x)\left[x(1-x)\right]^{n+1}},x=\frac{1}{2}\right)=-2\cdot 4^n. $$

$$ \begin{align} \sum_{k=0}^n2^k\binom{2n-k}{n} &=\sum_{k=0}^n\sum_{j=0}^k\binom{k}{j}\binom{2n-k}{n}\tag1\\ &=\sum_{j=0}^n\sum_{k=j}^n\binom{k}{j}\binom{2n-k}{n}\tag2\\ &=\sum_{j=0}^n\binom{2n+1}{n+j+1}\tag3\\ &=\sum_{j=0}^n\binom{2n+1}{n-j}\tag4\\ &=\frac12\sum_{j=0}^{2n+1}\binom{2n+1}{j}\tag5\\ &=\frac12\cdot2^{2n+1}\tag6\\[12pt] &=4^n\tag7 \end{align} $$ Explanation:
$(1)$: use the binomial theorem to expand $(1+1)^n$
$(2)$: change order of summation
$(3)$: Vandermonde Identity
$(4)$: symmetry of Pascal's Triangle
$(5)$: average $(3)$ and $(4)$
$(6)$: use the binomial theorem to expand $(1+1)^{2n+1}$
$(7)$: simplify

Step $(3)$ is actually an extension of Vandermonde using negative binomial coefficients: $$ \begin{align} \sum_{k=j}^n\binom{k}{j}\binom{2n-k}{n} &=\sum_{k=j}^n\binom{k}{k-j}\binom{2n-k}{n-k}\\ &=\sum_{k=j}^n(-1)^{k-j}\binom{-j-1}{k-j}(-1)^{n-k}\binom{-n-1}{n-k}\\ &=(-1)^{n-j}\binom{-n-j-2}{n-j}\\ &=\binom{2n+1}{n-j}\\ &=\binom{2n+1}{n+j+1} \end{align} $$

With formal power series as requested by OP we may write

$$\sum_{k=0}^n 2^k {2n-k\choose n} = 2^n \sum_{k=0}^n 2^{-k} {n+k\choose n} = 2^n \sum_{k=0}^n 2^{-k} [z^n] (1+z)^{n+k} \\ = 2^n [z^n] (1+z)^n \sum_{k=0}^n 2^{-k} (1+z)^k = 2^n [z^n] (1+z)^n \frac{1-(1+z)^{n+1}/2^{n+1}}{1-(1+z)/2} \\ = 2^{n+1} [z^n] (1+z)^n \frac{1-(1+z)^{n+1}/2^{n+1}}{1-z}.$$

We get from the first piece

$$2^{n+1} [z^n] (1+z)^n \frac{1}{1-z} = 2^{n+1} \sum_{k=0}^n {n\choose k} = 2^{2n+1}.$$

The second piece yields

$$- [z^n] (1+z)^{2n+1} \frac{1}{1-z} = - \sum_{k=0}^n {2n+1\choose k} = - \frac{1}{2} 2^{2n+1}.$$

Joining the two pieces we find

$$2^{2n+1} - 2^{2n} = 2^{2n} = 4^n.$$