Decomposing $C_{N_\gamma}$ amounts to finding all non-zero eigenvalues as well as corresponding eigenvectors: Assume for now that we already found all eigenvalues $\lambda_j>0$ as well as normalized eigenvectors $v_j$ (if they correspond to the same eigenvalue let's assume that we already made them orthogonal via Gram-Schmidt).
Then, because the matrix of interest is Hermitian we automatically obtain the decomposition $C_{N_{\gamma}}=\sum_j\lambda_j|v_j\rangle\langle v_j|$. Following the answers FDGod linked in their comment all we have to do from here is arrange the $v_j$ into matrices $V_j$ (via the inverse of the vectorization operation) which altogether yields Kraus operators $K_j:=\sqrt{\lambda_j}V_j$, as desired.
Now let us do some basic linear algebra to first determine the eigenvalues. Up to permutation (which does not change the
spectrum as it is a similarity transformation) our matrix is of the form
$$
\underbrace{\begin{pmatrix}
1&\sqrt{1-\gamma}\\
\sqrt{1-\gamma}&1-\gamma
\end{pmatrix}}_{=:X}\oplus(\gamma)\oplus(0)\,.
$$
Thus the eigenvalues of $C_{N_\gamma}$ are $\gamma,0,$ and whatever the eigenvalues of $X$ are. For the latter we compute
$$
\begin{align}
\det(\lambda\cdot{\bf1}-X)&=(\lambda-1)(\lambda-1+\gamma)-(1-\gamma)\\&=\lambda^2-\lambda+\gamma\lambda-\lambda\\&=\lambda(\lambda-2+\gamma)
\end{align}
$$
the roots of which are obviously $0$ and $2-\gamma$. (Note: A trained eye would see right away that $X$ is a rank-1 matrix so the only non-zero eigenvalue must be given by the trace). Either way this shows that the non-zero eigenvalues of $C_{N_\gamma}$ are $\gamma,2-\gamma$ whenever $\gamma\in(0,1]$, and $1$ if $\gamma=0$ (in which case $N_0$ is the identity so let us disregard this trivial case).
Next, the eigenvectors. For the eigenvalue $\gamma$, looking at $C_{N_\gamma}$ the corresponding (normalized) eigenvector is obviously $(0,0,1,0)^T$. For the second eigenvalue the problem boils down to finding the corresponding eigenvector for $X$, i.e. finding $v\in\mathbb C^2\setminus\{0\}$ such that
$$
0=((2-\gamma){\bf1}-X)v=\begin{pmatrix}
-\sqrt{1-\gamma}( -\sqrt{1-\gamma}\,v_1+v_2 )\\-\sqrt{1-\gamma}\,v_1+v_2
\end{pmatrix}\,.
$$
This linear system is solved, for example, by $v=(1,\sqrt{1-\gamma})^T$; again we could've taken a shortcut here by seeing that $X$ is rank-1 so $X=|v\rangle\langle v|$ where the entries of $v$ are necessarily the square root of the diagonal of $X$. Because $X$ "sits in" the first and last column of $C_{N_\gamma}$ the (normalized) eigenvector to the eigenvalue $2-\gamma$ reads $(1,0,0,\sqrt{1-\gamma})^T/\sqrt{2-\gamma}$.
All that is left to do now is to rearrange the eigenvectors into matrices $V_j$:
$$
\begin{pmatrix}0\\0\\1\\0\end{pmatrix}\to V_1=\begin{pmatrix}0&1\\0&0\end{pmatrix}\\\frac1{\sqrt{2-\gamma}}\begin{pmatrix}1\\0\\0\\\sqrt{1-\gamma}\end{pmatrix}\to V_2=\frac1{\sqrt{2-\gamma}}\begin{pmatrix}1&0\\0&\sqrt{1-\gamma}\end{pmatrix}
$$
This reproduces the well-known Kraus operators $K_j=\sqrt{\lambda_j}V_j$ of the amplitude damping channel:
$$
\boxed{K_1=\begin{pmatrix}0&\sqrt\gamma\\0&0\end{pmatrix}\qquad K_2=\begin{pmatrix}1&0\\0&\sqrt{1-\gamma}\end{pmatrix}}
$$
