Revised to include details of general $\alpha$
While not a complete answer, here are some ideas that may be of interest.
Define $\theta = (1+\alpha)^2$.
To ease notation, I'll write $K_L$ to mean $K_L(\alpha)$. Some experimentation reveals a nice pattern satisfied by $M_L := K_L^TK_L$. In particular, we see that $M_L$ is comprised of two numbers, $\frac{\theta}{2^L}$ and $\frac{2\theta}{2^L}$, which occur in a well-structured pattern. We seek to bound the Perron-Frobenius root, say $\rho$ of the nonnegative matrix $M_L$. Looking at the pattern of $M_L$ we see that the smallest possible column sum of $M_L$ happens when $\theta$ arises $2^{L-1}-1$ times and $2\theta$ arises once. Thus, we see that \begin{equation*} \rho \ge (2^{L-1}-1)\frac{\theta}{2^L} + \frac{2\theta}{2^L} = \theta\left(\frac{1}{2}+\frac{1}{2^L}\right). \end{equation*}
Similarly, by looking at the pattern, we see that the largest row-sum happens when $\theta$ appears $2^{L-2}$ times and $2\theta$ appears $2^{L-1}-2^{L-2}$ times (noting that $M_L$ is of size $2^{L-1} \times 2^{L-1}$). Thus, we see that \begin{equation*} \rho \le 2^{L-2}\frac{\theta}{2^L} + (2^{L-1}-2^{L-2}) \frac{2\theta}{2^L} = \frac{3}{4}\theta. \end{equation*}
Example, when $\alpha=0$, then $\theta=1$, and the upper bound on $\sigma_1(K_L) = \sqrt{\rho} \le \sqrt{3/4}$, otherwise, we get \begin{equation*} \sigma_1(K_L) \le (1+\alpha)\sqrt{\frac{3}{4} }. \end{equation*}
As I noted previously, a more careful analysis is needed to refine these bounds and to characterize how the limiting value is achieved (perhaps we can leverage $K_L$ being a column stochastic matrix?), but this bound is numerically not too bad. For example, a quick numerical experiment shows that $\sigma_1(K_L(0)) \to 0.8254....$ very rapidly (already for $K_7$, the first 4 digits match).