Proof for $\sum_{i=0}^n \frac{n \choose i}{{2n-1} \choose i} = 2$.

From the hockey stick identity we have \begin{eqnarray*} \sum_{i=0}^{n} \binom{2n-i-1}{n-1} = \binom{2n}{n} = 2 \binom{2n-1}{n}. \end{eqnarray*} This can be rearranged to give your identity.

Alternatively, using the Beta function \begin{eqnarray*} \binom{2n-1}{i} ^{-1} = 2n \int_0^1 t^i (1-t)^{2n-i-1} dt. \end{eqnarray*} Substituting this gives \begin{eqnarray*} \sum_{i=0}^{n} \binom{n}{i} \binom{2n-1}{i}^{-1} & = & 2n \int_0^1 \sum_{i=0}^{n} \binom{n}{i} t^i (1-t)^{2n-i-1} dt \\ & = & 2n \int_0^1 \left( 1+ \frac{t}{1-t} \right)^n (1-t)^{2n-1} dt \\ & = & 2n \int_0^1 (1-t)^{n-1} dt = \color{red}{2}. \\ \end{eqnarray*} And this method might be more useful when trying to generalise your identity.

In general, for any integers $n,k\ge 0$ are, we have
$$ \boxed{\sum_{i=0}^n \frac{\binom{n}{i}}{\binom{n+k}i}=\frac{n+k+1}{k+1}.} $$ Your problems is the special case $k=n-1$.

To see this, imagine you have a shuffled deck with $n$ black cards and $k$ red cards. You deal cards from the top until you get a red card. What is exepcted number of cards dealt?

Letting $X$ be the number of cards dealt, then $$ E[X]=\sum_{i=0}^\infty P(X>i)=\sum_{i=0}^n \frac{\binom{n}{i}}{\binom{n+k}i} $$ because the event $\{X>i\}$ occurs if the first $i$ cards are all chosen from the $n$ black cards, out of a total of $\binom{n+k}{i}$ ways to choose the first $i$ cards.

On the other hand, the $k$ red cards divide the deck into $k+1$ sections. Each black card is equally likely to fall in each section, so the probability a particular black card is in the top section is $\frac1{k+1}$. Therefore, the expected number of black cards in the top section is $n\cdot \frac1{k+1}$, so the expected number of cards dealt is $$E[X]=1+\frac{n}{k+1}=\frac{n+k+1}{k+1}.$$