We will only cover the case $K_0 = K_1 = R_0 = R_1 = 1$ here.
Let us look at sequence $K_n$ first. It turns out we can express $K_n$ in a closed form.
The key is following observation:
$$K_n = \frac{1}{K_{n-1}} + K_{n-2}
\quad\iff\quad K_nK_{n-1} - K_{n-1}K_{n-2} = 1$$
Summing over them leads to
$$K_nK_{n-1} = K_1K_0 + \sum_{k=2}^n (K_kK_{k-1}-K_{k-1}K_{k-2})
= 1 + (n-1) = n
$$
For even $n = 2k$, we have
$$K_n = K_{2k} = K_0 \prod_{j=1}^k \frac{K_{2j}K_{2j-1}}{K_{2j-1}K_{2j-2}}
= \prod_{j=1}^k \frac{2j}{2j-1} = \prod_{j=1}^k \frac{(2j)^2}{2j(2j-1)}
= \frac{2^{2k} (k!)^2}{(2k)!}.
$$
For odd $n = 2k+1$, we have
$$K_n = K_{2k+1} = \frac{K_{2k+1}K_{2k}}{K_{2k}} = \frac{(2k+1)(2k)!}{2^{2k}(k!)^2} = \frac{(2k+1)!}{2^{2k}(k!)^2}$$
For large even $n = 2k$, Stirling's approximation tell us
$$K_{2k} = \frac{2^{2k}\left(
\frac{k}{e}\right)^{2k}2\pi k}{\left(\frac{2k}{e}\right)^{2k}\sqrt{2\pi(2k)}}(1 + O(k^{-1}))
= \sqrt{\pi k} + O(k^{-1/2})
$$
Together with $K_{2k+1}K_{2k} = 2k+1$, we find for large $n$, $\displaystyle\;K_n^2 = \begin{cases}
\frac{\pi}{2} n, & n \text{ even }\\
\frac{2}{\pi} n, & n \text{ odd }
\end{cases} + O(1)$.
Let us switch to sequence $R_n$, the recurrence relation
$\displaystyle\;R_n = R_{n-1} + \frac{1}{R_{n-2}}$
tells us if $R_{n-1}, R_{n-2}$ is positive, so does $R_n$. Since $R_0 = R_1 = 1$ is positive, all $R_n$ is positive and $R_n$ is an increasing sequence.
Notice $R_2 = 2$ and for all $n > 2$,
$$R_n^2 = \left(R_{n-1} + \frac{1}{R_{n-2}}\right)^2
= R_{n-1}^2 + 2\frac{R_{n-1}}{R_{n-2}} + \frac{1}{R_{n-2}^2}
= R_{n-1}^2 + 2 + \frac{2}{R_{n-2}R_{n-3}} + \frac{1}{R_{n-2}^2}\tag{*1}
$$
We find for $n > 2$, we can bound $R_n^2$ from below as
$$R_n^2 = R_2^2 + \sum_{k=3}^n \left( R_k^2 - R_{k-1}^2 \right) \ge 4 + \sum_{k=3}^n 2 = 4+2(n-2) = 2n$$
Please note that this inequality $R_n^2 \ge 2n$ is also valid at $n = 2$.
Now substitute this lower bound back to $(*1)$, we find for $n > 4$
$$\begin{align} R_n^2 - 2n
&= \sum_{k=3}^n(R_k^2 - R_{k-1}^2 - 2)
= \sum_{k=3}^n \left(\frac{2}{R_{k-2}R_{k-3}} + \frac{1}{R_{k-2}^2}\right)\\
&\le
\sum_{k=3}^n \frac{3}{R_{k-3}^2}
\le 6 + \sum_{k=5}^n \frac{3}{2(k-3)}
= \frac{9 + 3H_{n-3}}{2}
\end{align}
$$
where $H_n = \sum_{k=1}^n \frac{1}{k}$ is the $n^{th}$ harmonic number.
For large $n$, we have $H_n \approx \log n$. As a result,
$$R_n^2 = 2n + O(\log n)$$
This leads to
$$\begin{align}
\lim_{k\to\infty}\frac{R_{2k}^2}{K_{2k}^2}
&= \lim_{k\to\infty}\frac{2(2k) + O(\log k)}{\frac{\pi}{2}(2k) + O(1)} = \frac{4}{\pi}\\
\lim_{k\to\infty}\frac{R_{2k+1}^2}{K_{2k+1}^2}
&=
\lim_{k\to\infty}\frac{2(2k+1) + O(\log k)}{\frac{2}{\pi}(2k+1) + O(1)} = \pi
\end{align}
$$
This is pretty close to what OP has got except the limits for odd and even case has been swapped. I have computed the sequences $R_n$ and $K_n$ myself and
their ratios agree with result here.
