To find the Kraus operators from the environmental form it is useful to introduce the operator $\iota_\phi:={\bf1}\otimes|\phi\rangle$
(with $|\phi\rangle\in\mathbb C^2$ is arbitrary; acts like $\iota_\phi(|x\rangle)=|x\rangle\otimes|\phi\rangle$) because then
- $A\otimes|\phi\rangle\langle\phi|=\iota_\phi A\iota_\phi^\dagger$ for all $A\in\mathbb C^{2\times 2}$, and
- ${\rm tr}_E(B)=\iota_0 B\iota_0^\dagger+\iota_1 B\iota_1^\dagger$ for all $B\in\mathbb C^2\otimes\mathbb C^2$.
This way, one finds that $\{E_0,E_1\}:=\{\iota_0^\dagger U\iota_\psi,\iota_1^\dagger U\iota_\psi\}$ are Kraus operators of ${\rm tr}_E(U((\cdot)\otimes|\psi\rangle\langle \psi|)U^\dagger)=E_0(\cdot)E_0^\dagger+E_1(\cdot)E_1^\dagger$. For more on this formalism refer to Chapter 8.2.3 in the book "Quantum Computation and Quantum Information" of Nielsen and Chuang (alt link).
To determine the matrix form of these Kraus operators we can just insert the matrix forms of $U$, $\iota$:
\begin{align*}
E_0&=\iota_0^\dagger U\iota_\psi=\big({\bf1}\otimes|0\rangle\big)^\dagger U\big({\bf1}\otimes |\psi\rangle\big)\\
&=\Big(\begin{pmatrix}1&0\\0&1\end{pmatrix}\otimes\begin{pmatrix}1&0\end{pmatrix}\Big)\begin{pmatrix}1&0&0&0\\ 0&\sqrt{1-\gamma}&\sqrt\gamma&0 \\ 0&-\sqrt\gamma&\sqrt{1-\gamma}&0 \\0&0&0&1 \end{pmatrix}\Big(\begin{pmatrix}1&0\\0&1\end{pmatrix}\otimes\begin{pmatrix}\sqrt{1-p}\\\sqrt p\end{pmatrix}\Big)\\
&=\begin{pmatrix}1&0&0&0\\0&0&1&0\end{pmatrix}\begin{pmatrix}1&0&0&0\\ 0&\sqrt{1-\gamma}&\sqrt\gamma&0 \\ 0&-\sqrt\gamma&\sqrt{1-\gamma}&0 \\0&0&0&1 \end{pmatrix}\begin{pmatrix}\sqrt{1-p}&0\\\sqrt p&0\\0&\sqrt{1-p}\\0&\sqrt p\end{pmatrix}\\
&=\begin{pmatrix}1&0&0&0\\0&0&1&0\end{pmatrix}\begin{pmatrix}
\sqrt{1-p}&0\\
\sqrt{(1-\gamma)p}&\sqrt{\gamma(1-p)}\\
-\sqrt{\gamma p}&\sqrt{(1-\gamma)(1-p)}\\
0&\sqrt p
\end{pmatrix}\\
&=\begin{pmatrix}
\sqrt{1-p}&0\\
-\sqrt{\gamma p}&\sqrt{(1-\gamma)(1-p)}
\end{pmatrix}
\end{align*}
Analogously one finds
$$
E_1=\begin{pmatrix}
\sqrt{(1-\gamma)p}&\sqrt{\gamma(1-p)}\\0&\sqrt p
\end{pmatrix}.
$$
