Simplifying we have:
$$S_k = \frac{\sum_{i=1}^k i\binom{k}{2i-k}\binom{2k - i - 1}{k - 1}}{k\binom{2k-1}{k}}.$$
It follows that $S_k$ equals the coefficient of $x^k$ in
$$\frac{1+2x}{2(1+x)(1-x)^kk\binom{2k-1}{k}}.$$
Using Lagrange inversion, we further have it as the coefficient of $x^k$ in $\frac{f(x)}{\binom{2k-1}{k}}$, where
$$f(x):=\frac{1+x+(1+2x)(1-4x)^{-1/2}}{2(2+x)}.$$
For $k\geq 1$, we further have
$$\frac1{k\binom{2k-1}{k}} = \int_0^1 t^{k-1}(1-t)^{k-1} dt$$
and thus
$$S_k = [x^{k-1}]\int_0^1 f'(xt(1-t))dt.$$
Computing this integral, we get the generating function:
$${\cal S}(x):=\sum_{k\geq 1} S_k x^k = \frac{x^{1/2}}{(8+x)^{3/2}}\left(4\operatorname{arctanh}\frac{x^{1/2}}{(8+x)^{1/2}} - 2\operatorname{arctanh}\frac{x-1-x^{1/2}(8+x)^{1/2}}3 + 2\operatorname{arctanh}\frac{x-1+x^{1/2}(8+x)^{1/2}}3\right) + \frac{2}{3(1-x)} - \frac{16}{3(8+x)}.$$

This function has poles at $x=1$ and $x=-8$ and thus $$\lim_{k\to\infty} S_k = \lim_{x\to 1}(1-x){\cal S}(x) = \frac23.$$