$\newcommand{\la}{\lambda}$Up to the equality in distribution, \begin{equation*} Y=\sqrt{\sum_1^k(Z_i+L_k)^2}, \end{equation*} where $Z,Z_1,\dots,Z_k$ are independent standard normal random variables (r.v.'s) and \begin{equation*} L_k:=\la/\sqrt k\to L \end{equation*} (as $k\to\infty$).
By the law of large numbers, \begin{equation*} \frac{Y^2}k=\frac1k\sum_1^k(Z_i+L)^2+(L_k-L)^2+2(L_k-L)\frac1k\sum_1^k (Z_i+L) \\ \to E(Z+L)^2=1+L^2 \end{equation*} and hence \begin{equation*} \frac{Y}{\sqrt k}\to\sqrt{1+L^2} \end{equation*} in probability and hence in distribution.
Also, $Var((Z+L_k)^2)=2+4L_k^2$ and hence $Var(Y^2)=k\,Var((Z+L_k)^2)=2k+4\la^2\sim2k(1+2L^2)$. So, by the central limit theorem (or, more precisely, by, say, the Berry--Esseen inequality), \begin{equation*} \frac{Y^2-(k+\la^2)}{\sqrt{2k(1+2L^2)}}\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{\sqrt{1+2L^2}}{\sqrt{2(1+L^2)}} \,Z \end{equation*} in distribution.
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{\sqrt{1+2L^2}}{\sqrt{2(1+L^2)}} \,Z=0 \end{equation*} and \begin{equation*} E(Y-\sqrt{k+\la^2})^2\to E\Big(\frac{\sqrt{1+2L^2}}{\sqrt{2(1+L^2)}} \,Z\Big)^2=\frac{{1+2L^2}}{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\frac{{1+2L^2}}{2(1+L^2)}. \tag{2}\label{2} \end{equation*}