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: $$\sum_{k\geq 1} S_k x^k = \frac{x^{1/2}}{(8+x)^{3/2}}\left(4\operatorname{arctanh}\frac{x^{1/2}}{(8+x)^{1/2}} - 2\operatorname{arctanh}\frac{x-1-x^{1/2}(8+x)^{1/2}}3 + 2\operatorname{arctanh}\frac{x-1+x^{1/2}(8+x)^{1/2}}3\right) + \frac{2}{3(1-x)} - \frac{16}{3(8+x)}.$$
This function has a poles at $x=1$ and $x=-8$ and thus $\lim_{k\to\infty} S_k = \frac23$.