How prove this $\sum_{k=0}^{n} \frac{\binom{2n-k}{n}}{2^{2n-k}}=1 $

Things may be more clear if we let $k=n-i$. Then our sum (reversed) is $$\sum_{i=0}^n \frac{\binom{n+i}{n}}{2^{n+i}}.$$ Imagine tossing a fair coin until we get $n+1$ heads or $n+1$ tails.

The number of tosses is $n+i+1$ if (i) we get exactly $n$ heads in the first $n+i$ tosses, and then a head, or (ii) we get $n$ tails in the first $n+i$ tosses and then a tail.

The probability of (i) is $\frac{\binom{n+i}{n}}{2^{n+i}}\cdot \frac{1}{2}$, and the probability of (ii) is the same. For sure the number of tosses will be one of $n+1$ to $2n+1$.

Take the set of paths from $0$ to $2n$, and break it up by the smallest $k$ where the path hits $|y|=2n-x$. This gives $$\sum_{k=0}^n {2n-k\choose n-k} 2^k=4^n,$$ which is the same as your formula.