$\newcommand{\B}{\vec{B}^\times}
\newcommand{\e}{\vec{E}}
\renewcommand{\b}{\vec{\beta}}
\newcommand{\bv}{\vec{B}}$
The field tensor can be written
$\begin{pmatrix}
0 & -\e \\
\e & \B
\end{pmatrix}$,
Where $\B$ is the dual tensor to $\vec{B}$ defined by $\B \vec{v} = \vec{B} \times \vec{v}$. Equivalently, $(\B)_{ik} = \epsilon_{ijk} B_j$. Note $\vec{v}^T \B = (\vec{v} \times \vec{B})^T$. It will also be important to note that $$(\vec{v} \times \vec{w})^\times \vec{u} = \vec{w} (\vec{u} \cdot \vec{v}) - \vec{v} (\vec{u} \cdot \vec{w})$$, so that $$ (\vec{v} \times \vec{w})^\times = \vec{w} \otimes \vec{v} - \vec{v} \otimes \vec{w}$$.
The action of a Lorentz transformation can be written
$$\begin{pmatrix}
\gamma & -\gamma \b \\
-\gamma \b& 1+\alpha \b \otimes \b
\end{pmatrix}$$, where $$\alpha = \frac{\gamma^2}{1+\gamma}$$. It will be important to note that $$\gamma^2 - \gamma \alpha = \gamma(\gamma - \alpha) = \gamma (\frac{\gamma + \gamma^2 - \gamma^2}{1 + \gamma}) = \frac{\gamma^2}{1+\gamma} = \alpha$$. Also $$1+\alpha \beta^2 = 1 + \alpha (1-1/\gamma^2) =1+ \frac{\gamma^2 -1}{1 + \gamma} =1+ \gamma -1 = \gamma$$
Anyway, the transformed field is
$$\begin{pmatrix}
\gamma & -\gamma \b \\
-\gamma \b& 1+\alpha \b \otimes \b
\end{pmatrix}
\begin{pmatrix}
0& -\e \\
\e& \B
\end{pmatrix}
\begin{pmatrix}
\gamma & -\gamma \b \\
-\gamma \b& 1+\alpha \b \otimes \b
\end{pmatrix}$$.
Since the field tensor is antisymmetric, and the Lorentz transformation tensor is symmetric, we know the result must be antisymmetric. We will use this fact later. Let's start by compute the first product
$$\begin{pmatrix}
\gamma & -\gamma \b \\
-\gamma \b& 1+\alpha \b \otimes \b
\end{pmatrix}
\begin{pmatrix}
0& -\e \\
\e& \B
\end{pmatrix} =
\begin{pmatrix}
-\gamma \b \cdot \e& -\gamma \e-\gamma \b \times \bv \\
\e + \alpha \b (\b \cdot \e)& \gamma \b \otimes \e + \B + \alpha \b \otimes (\b \times \bv)
\end{pmatrix}
$$.
Next we compute the second product. Since we already know this product will be antisymmetric, we will only calculate the right column.
$$
\begin{pmatrix}
-\gamma \b \cdot \e& -\gamma \e-\gamma \b \times \bv \\
\e + \alpha \b (\b \cdot \e)& \gamma \b \otimes \e + \B + \alpha \b \otimes (\b \times \bv)
\end{pmatrix}
\begin{pmatrix}
\gamma & -\gamma \b \\
-\gamma \b& 1+\alpha \b \otimes \b
\end{pmatrix}
$$
$$
= \begin{pmatrix}
0 & \gamma^2 \b (\b \cdot \e) -\gamma \e-\gamma \b \times \bv -\alpha \gamma \b (\b \cdot \e) \\
\cdots & -\gamma \e \otimes \b - \alpha \gamma (\b \cdot \e) \b \otimes \b + \gamma \b \otimes \e + \B \\
& + \alpha \b \otimes (\b \times \bv) + \alpha \gamma (\b \cdot \e) \b \otimes \b + \alpha (\bv \times \b) \otimes \b
\end{pmatrix}
$$
$$
= \begin{pmatrix}
0 & -(\gamma (\e + \b \times \bv) - (\gamma^2 - \alpha \gamma) \b (\b \cdot \e)) \\
\cdots & \B -\gamma( \e \otimes \b - \b \otimes \e) - \alpha((\b \times \bv) \otimes \b - \b \otimes (\b \times \bv))
\end{pmatrix}
$$
$$
=\begin{pmatrix}
0 & -(\gamma (\e + \b \times \bv) - \alpha \b (\b \cdot \e)) \\
\cdots & \B -\gamma( \b \times \e)^\times - \alpha(\b \times (\b \times \bv))^\times
\end{pmatrix}
$$
By now we have found the expected expression for the new electric field from the upper right entry: $\tilde{\e} =\gamma (\e + \b \times \bv) - \alpha \b (\b \cdot \e)$. Let's now focus on the bottom right entry.
$$ \tilde{\bv}^\times=\B -\gamma( \b \times \e)^\times - \alpha((\b \cdot \bv) \b^\times - \beta^2 \B)$$
$$ = ((1+\alpha \beta^2)\bv - \gamma \b \times \e - \alpha \b (\b \cdot \bv) )^\times$$.
Thus $$\tilde{\bv} = \gamma \bv - \gamma \b \times \e - \alpha \b (\b \cdot \bv) $$
$$ = \gamma(\bv - \b \times \e) - \alpha \b (\b \cdot \bv)$$ as was desired.
