Tramsformation of spatial components of the 4-force

Question

I'm trying to learn special relativity by myself. I've been reading the Griffith's chapter about relativistic dynamics and electrodynamics (chapter 12), but one thing it's not clear to me. I've been trying to derive the formula $12.67$:

for a particle instantaneously at rest in a reference frame $S$, such that $\vec{u} = \vec{0}$:

$$ \vec{F'_\perp} = \frac{1}{\gamma} \vec{F_\perp}$$ $$ \vec{F'_\parallel} = \vec{F_\parallel} $$

where $\gamma$ is referred to the motion of of an inertial frame $S'$ w.r.t $S$.

I've tried to use two different methods than the one proposed by the author, but none of them seems to work:

1. First method. I know that, if the mass of the object $m_0$ is constant

$$ F^\mu = \frac{dp^\mu}{d\tau} = m_0 \frac{d v^\mu}{d\tau} $$

(for convenience I will write contravariant vectors as row vectors)

$$ \frac{d v^\mu}{d \tau} = \frac{d v^\mu}{dt} \frac{dt}{d\tau} = \gamma \frac{d}{dt} \left( \gamma \left(c, \vec{v}\right) \right) = \gamma \left( \frac{d\gamma}{dt}\left( c, \vec{v} \right) + \gamma \frac{d}{dt} \left( c, \vec{v} \right) \right) = \gamma \left( \frac{d \gamma}{dt} \left(c, \vec{v} \right) + \gamma \left(0, \vec{a} \right) \right)$$

$$ \frac{d\gamma}{dt} = \frac{d}{dt} \frac{1}{\sqrt{1-v^2/c^2}} = -\frac{1}{2} \left( 1 - v^2/c^2 \right)^{-3/2} \frac{1}{c^2} \frac{d}{dt} \left(-v^2\right) = -\frac{1}{2c^2} \gamma^3 \frac{d}{dt}\left(\left( 0, \vec{v} \right) \cdot \left( 0, \vec{v} \right) \right) = -\frac{1}{2c^2} \gamma^3 2\vec{v} \cdot \vec{a} = -\frac{\vec{v} \cdot \vec{a}}{c^2} \gamma^3$$

Putting it all together:

$$ \frac{dv^\mu}{d\tau} = -\gamma^4 \frac{\vec{v} \cdot \vec{a}}{c^2} v^\mu + \gamma^2\left(0, \vec{a}\right) $$

$$ F^\mu = -\gamma^4 m_0 \frac{\vec{v} \cdot \vec{a}}{c^2} v^\mu + \gamma^2 m_0 \left(0, \vec{a}\right) $$

Considering only the spatial components of the force 4-force:

$$\vec{F} = -\gamma^4 m_0 \frac{\vec{v} \cdot \vec{a}}{c^2} \vec{v} + \gamma^2 m_0 \vec{a} = -\gamma^4 m_0 \frac{\vec{v}^T \vec{a}}{c^2}\vec{v} + \gamma^2m_0 \vec{a} = \gamma^2 m_0 a \left( -\gamma^2 \frac{\vec{v} \cdot \vec{v}}{c^2} + 1 \right) = \gamma^2 m_0 a \left( -\gamma^2 \beta^2 + 1 \right)$$

Now, $m_0 a$ is the classical expression for the force, but I can't figure out how to derive from this expression of the force the relation that the author is talking about. One would need to express $\vec{F}$ in function of $\vec{F'}$ or viceversa to try to obtain the relations of the author

2. Second method. The 4-force is a 4-vector, then:

$$ F'^\mu = \Lambda^\mu_\nu F^\nu \Longrightarrow \begin{bmatrix} F'^0 \\ \vec{F'} \end{bmatrix} = \Lambda \begin{bmatrix} F^0 & \\ \vec{F} \end{bmatrix}$$

The Lorentz boost in a generic direction has the form

$$ \Lambda = \begin{bmatrix} \gamma & -\gamma \vec{\beta^T} \\ -\gamma \vec{\beta} & Id_3 + \left( \gamma - 1 \right) \frac{\vec{\beta}\vec{\beta}^T}{\beta^2} \end{bmatrix} $$

Where $\vec{\beta} = \left( \beta_x, \beta_y, \beta_z \right)^T$. Then the spatial components of the 4-force are:

$$ \vec{F'} = -\gamma \vec{\beta} F^0 + \left[ Id_3 + \left( \gamma - 1 \right) \frac{\vec{\beta} \vec{\beta}^T}{\beta^2} \right] \vec{F} $$

In this case it shouldn't even be possible to set $\vec{v} = 0$, because of the $\beta^2$ at the denominator

This methods are most likely wrong, but I can't get where. Nonetheless, is there a way to derive the relations $12.67$ using 4-force?