$\newcommand{\la}{\lambda}$Write \begin{equation*} Y=\sqrt{V_k+\la^2}, \end{equation*} where $V_k$ is a random variable (r.v.) with the chi-squared distribution with $k$ degrees of freedom.
So, by the law of large numbers and the condition $\la\sim L\sqrt k$ (as $k\to\infty$), \begin{equation*} \frac{Y}{\sqrt k}=\sqrt{\frac{V_k}k+\frac{\la^2}k}\to\sqrt{1+L^2} \end{equation*} in probability and hence in distribution. Also, by the central limit theorem (or, more precisely, by, say, the Berry--Esseen inequality), \begin{equation*} \frac{Y^2-(k+\la^2)}{\sqrt{2k}}\to Z\sim N(0,1) \end{equation*} in distribution. So, by Slutsky's theorem, \begin{equation*} Y-\sqrt{k+\la^2}=R_k:=\frac{(Y^2-(k+\la^2))/\sqrt{2k}}{(Y+\sqrt{k+\la^2})/\sqrt{2k}} \to\frac Z{\sqrt{2(1+L^2)}} \end{equation*} in distribution, where again $Z\sim N(0,1)$.
Also, by (say) the Rosenthal inequality, \begin{equation*} ER_k^4\le E\Big(\frac{(Y^2-(k+\la^2))/\sqrt{2k}}{(\sqrt{k+\la^2})/\sqrt{2k}}\Big)^4=O(1). \end{equation*} So, by uniform integrability, \begin{equation*} E(Y-\sqrt{k+\la^2})\to E\frac Z{\sqrt{2(1+L^2)}}=0 \end{equation*} and \begin{equation*} E(Y-\sqrt{k+\la^2})^2\to E\Big(\frac Z{\sqrt{2(1+L^2)}}\Big)^2=\frac1{2(1+L^2)}. \end{equation*} Thus, \begin{equation*} EY=\sqrt{k+\la^2}+o(1) \tag{1}\label{1} \end{equation*} and \begin{equation*} Var\, Y= E(Y-\sqrt{k+\la^2})^2-(E(Y-\sqrt{k+\la^2}))^2\to\frac1{2(1+L^2)}. \tag{2}\label{2} \end{equation*}
For an illustration of \eqref{1} and \eqref{2}, shown below are (parts of) the (connected) graphs $\{(k,EY-\sqrt{k+\la^2})\colon k\in\{1,\dots,50\}\}$ (left) and $\{(k,Var\,Y-\frac1{2(1+L^2)}\colon k\in\{1,\dots,50\}\}$ (right) for $\la=L\sqrt k$ with $L=0$ (red), $L=1$ (green), and $L=2$ (blue):
