I think the clearest way to think about this is to say that the gamma matrices don't transform. In other words, the fact that they carry a vector index doesn't mean that they form a four vector. This is analogous to how the Pauli matrices work in regular quantum mechanics, so let me talk a little bit about that.
Suppose you have a spin $1/2$ particle in some state $|\psi\rangle$. You can calculate the mean value of $\sigma_x$ by doing $\langle \psi | \sigma_x | \psi\rangle$. Now let's say you rotate your particle by an angle $\theta$ around the $z$-axis. (Warning: There is about a 50% chance my signs are incorrect.) You now describe your particle with a different ket, given by $|\psi'\rangle = \exp(-i \sigma_z \theta /2)$. Remember that we are leaving the coordinates fixed and rotating the system, as is usually done in quantum mechanics. Now the expectation value is given by
$$\langle \psi' | \sigma_x | \psi' \rangle = \langle \psi |\, e^{i\sigma_z \theta /2}\, \sigma_x\, e^{-i \sigma_z \theta / 2}\, | \psi\rangle$$
There is a neat theorem, not too hard to prove, that says that
$$e^{i\sigma_z \theta /2}\, \sigma_x\, e^{-i \sigma_z \theta / 2} = \cos \theta\, \sigma_x -\sin \theta\, \sigma_y$$
So it turns out that the expectation value for the rotated system is also given by $\langle \psi |\, \cos \theta\, \sigma_x -\sin \theta\, \sigma_y \, |\psi\rangle = \cos \theta\, \langle \sigma_x \rangle - \sin \theta\, \langle \sigma_y \rangle$. It's as if we left our particle alone and rotated the Pauli matrices. But note that if we apply the rotation to $|\psi\rangle$, then we don't touch the matrices. Also, I never said that I transformed the matrices. I just transformed the state, and then found out that I could leave it alone and rotate the matrices.
The situation for a Dirac spinor is similar. The analogous identity is that $S(\Lambda) \gamma^\mu S^{-1}(\Lambda) \Lambda^\nu_{\ \mu} = \gamma^\nu$. This is just something that is true; nobody said anything about transforming $\gamma^\mu$. There's no $\gamma^\mu \to \dots$ here.
Now let's take the Dirac equation, $(i \gamma^\mu \partial_\mu - m)\psi = 0$, and apply a Lorentz transformation. This time I will change coordinates instead of boosting the system, but there's no real difference. Let's say we have new coordinates given by $x'^\mu = \Lambda^\mu_{\ \nu} x^\nu$, and we want to see if the Dirac equation looks the same in those coordinates. The field $\psi'$ as seen in the $x'^\mu$ frame is given by $\psi' = S(\Lambda) \psi \iff \psi = S^{-1}(\Lambda) \psi'$, and the derivatives are related by $\partial_\mu = \Lambda^\nu_{\ \mu} \partial'_\nu$. Plugging in we get $(i\gamma^\mu \Lambda^\nu_{\ \mu} \partial'_\nu-m) S^{-1}(\Lambda)\psi' = 0$, which doesn't really look like our original equation. But let's multiply on the left by $S(\Lambda)$. $m$ is a scalar so $S$ goes right through it and cancels with $S^{-1}$. And in the first term we get $S(\Lambda)\gamma^\mu S^{-1}(\Lambda) \Lambda^\nu_{\ \mu}$, which according to our trusty identity is just $\gamma^\nu$. Our equation then simplifies to
$$(i\gamma^\mu \partial'_\mu - m)\psi'=0$$
This is the same equation, but written in the primed frame. Notice how the gamma matrices are the same as before; when you're in class and the teacher writes them on the board, you don't need to ask in what coordinate system they are valid. Everyone uses the same gamma matrices. They're not really a four-vector, but their "transformation law" guarantees that anything written as if they were a four vector is Lorentz invariant as long as the appropiate spinors are present.
The Pauli matrices are invariant tensors that couple left and right-handed spinors. These spinors transform in different representations of the Lorentz group (as you mentioned) and hence are usually denoted with different indices. This is trivial to see in two component notation, however if you are not familiar with this notation this can also be seen from a four-component Lagrangian:
$$
\bar{\psi} \gamma _\mu \psi = \psi^\dagger\gamma^0\gamma^\mu\psi=\left( \begin{array}{cc}\psi_R^* & \psi_L^* \end{array} \right) \left( \begin{array}{cc}
0 & \sigma ^\mu \\
\bar{\sigma} ^\mu & 0
\end{array} \right) \left( \begin{array}{c}
\psi _L \\
\psi _R
\end{array} \right) = \psi_R^*\sigma^\mu\psi_R +\psi_L^* \bar{\sigma} ^\mu \psi _L.
$$
One can then show that $\psi_L^*$ ($\psi_R^*$) transforms as a right-handed (left-handed) spinor. Clearly a $ \sigma ^\mu $ field then connects a $ \psi _R $ field with a $ \psi _L $ field. We can write these contractions more explicitly by denoting the left-handed representation indices by greek indices and right-handed reprentation indices with dotted greek indices:
$$
\psi^*_{L \, \dot{\alpha}} \left(\sigma^\mu\right)^{\dot{\alpha}}_{\phantom {\alpha}\alpha} \psi_L ^\alpha
$$
Note: you might be tempted to think of $\psi _R $ and $\psi_L$ not as separate fields, but just fields with projectors acting on them. This makes this whole topic very confusing and I would urge to get comfortable thinking in terms of two component fields as the fundamental objects making up fermions.
Best Answer
The point you are missing is that the bi-spinor $A_{\alpha\dot\alpha}$ and a vector $B_{m}$ are equivalent representations of $SO(1,3)$, so it is possible to construct an object $\sigma^{m}_{\alpha\dot\alpha}$ that is invariant under $SO(1,3)$. You can use this object to translated one representation into another by:
$$ A_{\alpha\dot\alpha} = A_{m}\sigma^{m}_{\alpha\dot\alpha}\qquad A_{m}=\frac{1}{2}\bar\sigma_{m}^{\dot\alpha\alpha}A_{\alpha\dot\alpha} $$
where $\bar\sigma_{m}^{\dot\alpha\alpha}=\varepsilon^{\alpha\beta}\varepsilon^{\dot\alpha\dot\beta}\eta_{mn}\sigma^{n}_{\beta\dot\beta}$. The same can be done for the $B$'s. You can check that this equations are right by transforming both sides in the equalities. Now, sometimes it is convenient to pretend that just one type of index in $\sigma^{m}_{\alpha\dot\beta}$ transform, such that $\sigma^{m}_{\alpha\dot\beta}$ will behave as a vector if we transform only $m$. If you turn transform just the spinorial indices than $\sigma^{m}_{\alpha\dot\alpha}$ will also transform as a vector but with the the inverse of the transformation since the combination of the transform in all indices should cancel.
Indeed an infinitesimal transformation on the spinorial indices is
$$ \delta (\sigma^{m}_{\alpha\dot\alpha}) =\Theta_{np}(\sigma^{np})_{\alpha}\,^{\beta}\sigma^{m}_{\beta\dot\alpha}+\Theta_{np}(\sigma^{np})^{\dot\beta}\,_{\dot\alpha}\sigma^{m}_{\alpha\dot\beta} = -\Theta^{m}\,_{n}\sigma^{n}_{\alpha\dot\alpha} $$