Here is a mathematical derivation. We use the sign convention $(+,-,-,-)$ for the Minkowski metric $\eta_{\mu\nu}$.
I) First recall the fact that
$SL(2,\mathbb{C})$ is (the double cover of) the restricted Lorentz group $SO^+(1,3;\mathbb{R})$.
This follows partly because:
There is a bijective isometry from the Minkowski space $(\mathbb{R}^{1,3},||\cdot||^2)$ to the space of $2\times2 $ Hermitian matrices $(u(2),\det(\cdot))$,
$$\mathbb{R}^{1,3} ~\cong ~ u(2)
~:=~\{\sigma\in {\rm Mat}_{2\times 2}(\mathbb{C}) \mid \sigma^{\dagger}=\sigma \}
~=~ {\rm span}_{\mathbb{R}} \{\sigma_{\mu} \mid \mu=0,1,2,3\}, $$
$$\mathbb{R}^{1,3}~\ni~\tilde{x}~=~(x^0,x^1,x^2,x^3) \quad\mapsto \quad\sigma~=~x^{\mu}\sigma_{\mu}~\in~ u(2), $$
$$ ||\tilde{x}||^2 ~=~x^{\mu} \eta_{\mu\nu}x^{\nu} ~=~\det(\sigma), \qquad \sigma_{0}~:=~{\bf 1}_{2 \times 2}.\tag{1}$$
There is a group action $\rho: SL(2,\mathbb{C})\times u(2) \to u(2)$ given by
$$g\quad \mapsto\quad\rho(g)\sigma~:= ~g\sigma g^{\dagger},
\qquad g\in SL(2,\mathbb{C}),\qquad\sigma\in u(2), \tag{2}$$
which is length preserving, i.e. $g$ is a pseudo-orthogonal (or Lorentz) transformation.
In other words, there is a Lie group homomorphism
$$\rho: SL(2,\mathbb{C}) \quad\to\quad O(u(2),\mathbb{R})~\cong~ O(1,3;\mathbb{R}) .\tag{3}$$
Since $\rho$ is a continuous map and $SL(2,\mathbb{C})$ is a connected set, the image $\rho(SL(2,\mathbb{C}))$ must again be a connected set. In fact, one may show so there is a surjective Lie group homomorphism$^1$
$$\rho: SL(2,\mathbb{C}) \quad\to\quad SO^+(u(2),\mathbb{R})~\cong~ SO^+(1,3;\mathbb{R}) , $$
$$\rho(\pm {\bf 1}_{2 \times 2})~=~{\bf 1}_{u(2)}.\tag{4}$$
The Lie group $SL(2,\mathbb{C})=\pm e^{sl(2,\mathbb{C})}$ has Lie algebra
$$ sl(2,\mathbb{C})
~=~ \{\tau\in{\rm Mat}_{2\times 2}(\mathbb{C}) \mid {\rm tr}(\tau)~=~0 \}
~=~{\rm span}_{\mathbb{C}} \{\sigma_{i} \mid i=1,2,3\}.\tag{5}$$
The Lie group homomorphism $\rho: SL(2,\mathbb{C}) \to O(u(2),\mathbb{R})$ induces a Lie algebra homomorphism
$$\rho: sl(2,\mathbb{C})\to o(u(2),\mathbb{R})\tag{6}$$
given by
$$ \rho(\tau)\sigma ~=~ \tau \sigma +\sigma \tau^{\dagger},
\qquad \tau\in sl(2,\mathbb{C}),\qquad\sigma\in u(2), $$
$$ \rho(\tau) ~=~ L_{\tau} +R_{\tau^{\dagger}},\tag{7}$$
where we have defined left and right multiplication of $2\times 2$ matrices
$$L_{\sigma}(\tau)~:=~\sigma \tau~=:~ R_{\tau}(\sigma),
\qquad \sigma,\tau ~\in~ {\rm Mat}_{2\times 2}(\mathbb{C}).\tag{8}$$
II) Note that the Lorentz Lie algebra $so(1,3;\mathbb{R}) \cong sl(2,\mathbb{C})$ does not$^2$ contain two perpendicular copies of, say, the real Lie algebra $su(2)$ or $sl(2,\mathbb{R})$. For comparison and completeness, let us mention that for other signatures in $4$ dimensions, one has
$$SO(4;\mathbb{R})~\cong~[SU(2)\times SU(2)]/\mathbb{Z}_2,
\qquad\text{(compact form)}\tag{9}$$
$$SO^+(2,2;\mathbb{R})~\cong~[SL(2,\mathbb{R})\times SL(2,\mathbb{R})]/\mathbb{Z}_2.\qquad\text{(split form)}\tag{10}$$
The compact form (9) has a nice proof using quaternions
$$(\mathbb{R}^4,||\cdot||^2) ~\cong~ (\mathbb{H},|\cdot|^2)\quad\text{and}\quad SU(2)~\cong~ U(1,\mathbb{H}),\tag{11}$$
see also this Math.SE post and this Phys.SE post. The split form (10) uses a bijective isometry
$$(\mathbb{R}^{2,2},||\cdot||^2) ~\cong~({\rm Mat}_{2\times 2}(\mathbb{R}),\det(\cdot)).\tag{12}$$
To decompose Minkowski space into left- and right-handed Weyl spinor representations, one must go to the complexification, i.e. one must use the fact that
$SL(2,\mathbb{C})\times SL(2,\mathbb{C})$ is (the double cover of) the complexified proper Lorentz group $SO(1,3;\mathbb{C})$.
Note that Refs. 1-2 do not discuss complexification$^2$. One can more or less repeat the construction from section I with the real numbers $\mathbb{R}$ replaced by complex numbers $\mathbb{C}$, however with some important caveats.
There is a bijective isometry from the complexified Minkowski space $(\mathbb{C}^{1,3},||\cdot||^2)$ to the space of $2\times2 $ matrices $({\rm Mat}_{2\times 2}(\mathbb{C}),\det(\cdot))$,
$$\mathbb{C}^{1,3} ~\cong ~ {\rm Mat}_{2\times 2}(\mathbb{C})
~=~ {\rm span}_{\mathbb{C}} \{\sigma_{\mu} \mid \mu=0,1,2,3\}, $$
$$ M(1,3;\mathbb{C})~\ni~\tilde{x}~=~(x^0,x^1,x^2,x^3) \quad\mapsto \quad\sigma~=~x^{\mu}\sigma_{\mu}~\in~ {\rm Mat}_{2\times 2}(\mathbb{C}) , $$
$$ ||\tilde{x}||^2 ~=~x^{\mu} \eta_{\mu\nu}x^{\nu} ~=~\det(\sigma).\tag{13}$$
Note that forms are taken to be bilinear rather than sesquilinear.
There is a surjective Lie group homomorphism$^3$
$$\rho: SL(2,\mathbb{C}) \times SL(2,\mathbb{C}) \quad\to\quad
SO({\rm Mat}_{2\times 2}(\mathbb{C}),\mathbb{C})~\cong~ SO(1,3;\mathbb{C})\tag{14}$$
given by
$$(g_L, g_R)\quad \mapsto\quad\rho(g_L, g_R)\sigma~:= ~g_L\sigma g^{\dagger}_R, $$
$$ g_L, g_R\in SL(2,\mathbb{C}),\qquad\sigma~\in~ {\rm Mat}_{2\times 2}(\mathbb{C}).\tag{15} $$
The Lie group
$SL(2,\mathbb{C})\times SL(2,\mathbb{C})$
has Lie algebra $sl(2,\mathbb{C})\oplus sl(2,\mathbb{C})$.
The Lie group homomorphism
$$\rho: SL(2,\mathbb{C})\times SL(2,\mathbb{C})
\quad\to\quad SO({\rm Mat}_{2\times 2}(\mathbb{C}),\mathbb{C})\tag{16}$$
induces a Lie algebra homomorphism
$$\rho: sl(2,\mathbb{C})\oplus sl(2,\mathbb{C})\quad\to\quad
so({\rm Mat}_{2\times 2}(\mathbb{C}),\mathbb{C})\tag{17}$$
given by
$$ \rho(\tau_L\oplus\tau_R)\sigma ~=~ \tau_L \sigma +\sigma \tau^{\dagger}_R,
\qquad \tau_L,\tau_R\in sl(2,\mathbb{C}),\qquad
\sigma\in {\rm Mat}_{2\times 2}(\mathbb{C}), $$
$$ \rho(\tau_L\oplus\tau_R) ~=~ L_{\tau_L} +R_{\tau^{\dagger}_R}.\tag{18}$$
The left action (acting from left on a two-dimensional complex column vector) yields by definition the (left-handed Weyl) spinor representation $(\frac{1}{2},0)$, while the right action (acting from right on a two-dimensional complex row vector) yields by definition the right-handed Weyl/complex conjugate spinor representation $(0,\frac{1}{2})$. The above shows that
The complexified Minkowski space $\mathbb{C}^{1,3}$ is a $(\frac{1}{2},\frac{1}{2})$ representation of the Lie group $SL(2,\mathbb{C}) \times SL(2,\mathbb{C})$, whose action respects the Minkowski metric.
References:
Anthony Zee, Quantum Field Theory in a Nutshell, 1st edition, 2003.
Anthony Zee, Quantum Field Theory in a Nutshell, 2nd edition, 2010.
$^1$ It is easy to check that it is not possible to describe discrete Lorentz transformations, such as, e.g. parity $P$, time-reversal $T$, or $PT$ with a group element $g\in GL(2,\mathbb{C})$ and formula (2).
$^2$ For a laugh, check out the (in several ways) wrong second sentence on p.113 in Ref. 1: "The mathematically sophisticated say that the algebra $SO(3,1)$ is isomorphic to $SU(2)\otimes SU(2)$." The corrected statement would e.g. be "The mathematically sophisticated say that the group $SO(3,1;\mathbb{C})$ is locally isomorphic to $SL(2,\mathbb{C})\times SL(2,\mathbb{C})$." Nevertheless, let me rush to add that Zee's book is overall a very nice book. In Ref. 2, the above sentence is removed, and a subsection called "More on $SO(4)$, $SO(3,1)$, and $SO(2,2)$" is added on page 531-532.
$^3$ It is not possible to mimic an improper Lorentz transformations $\Lambda\in O(1,3;\mathbb{C})$ [i.e. with negative determinant $\det (\Lambda)=-1$] with the help of two matrices $g_L, g_R\in GL(2,\mathbb{C})$ in formula (15); such as, e.g., the spatial parity transformation
$$P:~~(x^0,x^1,x^2,x^3) ~\mapsto~ (x^0,-x^1,-x^2,-x^3).\tag{19}$$
Similarly, the Weyl spinor representations are representations of (the double cover of) $SO(1,3;\mathbb{C})$ but not of (the double cover of) $O(1,3;\mathbb{C})$. E.g. the spatial parity transformation (19) intertwine between left-handed and right-handed Weyl spinor representations.
Best Answer
First let's recall how to construct the finite dimensional irreducible representations of the Lorentz group. Say $J_i$ are the three rotation generators and $K_i$ are the three boost generators. \begin{align*} L_x = &\begin{pmatrix} 0&0&0&0 \\ 0&0&0&0 \\ 0&0&0&-1 \\ 0&0&1&0 \end{pmatrix}& L_y = &\begin{pmatrix} 0&0&0&0 \\ 0&0&0&1 \\ 0&0&0&0 \\ 0&-1&0&0 \end{pmatrix}& L_z = &\begin{pmatrix} 0&0&0&0 \\ 0&0&-1&0 \\ 0&1&0&0 \\ 0&0&0&0 \end{pmatrix}\\ K_x = &\begin{pmatrix} 0&1&0&0 \\ 1&0&0&0 \\ 0&0&0&0 \\ 0&0&0&0 \end{pmatrix}& K_y = &\begin{pmatrix} 0&0&1&0 \\ 0&0&0&0 \\ 1&0&0&0 \\ 0&0&0&0 \end{pmatrix}& K_z = &\begin{pmatrix} 0&0&0&1 \\ 0&0&0&0 \\ 0&0&0&0 \\ 1&0&0&0 \end{pmatrix}\\ \end{align*} They satisfy $$ [J_i, J_j] = \varepsilon_{ijk} J_k \hspace{1 cm} [K_i, K_j] = -\varepsilon_{ijk} J_k \hspace{1 cm} [J_i, K_j] = \varepsilon_{ijk}K_k. $$ (Note that I am using the skew-adjoint convention for Lie algebra elements where I did not multiply by $i$.)
We then define $$ A_i = \frac{1}{2} (J_i - i K_i) \hspace{2 cm} B_i = \frac{1}{2}(J_i + i K_i) $$ which satisfy the commutation relations $$ [A_i, A_j] = \varepsilon_{ijk} A_k \hspace{2cm} [B_i, B_j] = \varepsilon_{ijk} B_k \hspace{2cm} [A_i, B_j] = 0. $$
Here is how you construct the representation of the Lorentz group: first, choose two non-negative half integers $j_1$ and $j_2$. These correspond to two spin $j$ representations of $\mathfrak{su}(2)$, which I will label $$ \pi'_{j}. $$ Recall that that $$ \mathfrak{su}(2) = \mathrm{span}_\mathbb{R} \{ -\tfrac{i}{2} \sigma_x, -\tfrac{i}{2} \sigma_y, -\tfrac{i}{2} \sigma_z \} $$ where $$ [-\tfrac{i}{2} \sigma_i, -\tfrac{i}{2} \sigma_j] = -\tfrac{i}{2}\varepsilon_{ijk} \sigma_k. $$ For this question, we only need to know the spin $1/2$ representation of $\mathfrak{su}(2)$, which is given by $$ \pi'_{\tfrac{1}{2}}( -\tfrac{i}{2}\sigma_i) = -\tfrac{i}{2} \sigma_i. $$
So okay, how do we construct the $(j_1, j_2)$ representation of the Lorentz group? Any Lie algebra element $X \in \mathfrak{so}(1,3)$ can written as a linear combination of $A_i$ and $B_i$: $$ X = \sum_{i = 1}^3 (\alpha_i A_i + \beta_i B_i). $$ (Note that we are actually dealing with the complexified version of the Lie algebra $\mathfrak{so}(1,3)$ because our definitions of $A_i$ and $B_i$ have factors of $i$, so $\alpha, \beta \in \mathbb{C}$.)
$A_i$ and $B_i$ form their own independent $\mathfrak{su}(2)$ algebras.
The Lie algebra representation $\pi'_{(j_1, j_2)}$ is then given by \begin{align*} \pi'_{(j_1, j_2)}(X) &= \pi'_{(j_1, j_2)}(\alpha_i A_i + \beta_j B_i) \\ &\equiv \pi'_{j_1}(\alpha_i A_i) \otimes \big( \pi'_{j_2}(\beta_j B_i) \big)^* \end{align*} where the star denotes complex conjugation.
Sometimes people forget to mention that you have to include the complex conjugation, but it won't work otherwise!
If $j_1 = 1/2$ and $j_2 = 1/2$, we have \begin{equation*} \pi_{\frac{1}{2}}'(A_i) = -\frac{i}{2}\sigma_i \otimes I \hspace{2cm} \big(\pi_{\frac{1}{2}}' (B_i)\big)^* = \frac{i}{2} I \otimes\sigma_i^*. \end{equation*} We can explicitly write out these tensor products in terms of a $2 \times 2 = 4$ dimensional basis. (Here I am using the so-called "Kronecker Product" to do this. That just a fancy name for multiplying all the elements of two $2\times 2$ cell-wise to get a $4 \times 4$ matrix.) \begin{align*} \pi_{(\frac{1}{2},\frac{1}{2})}'(A_x) &= -\frac{i}{2}\begin{pmatrix} 0&0&1&0 \\ 0&0&0&1 \\ 1&0&0&0 \\ 0&1&0&0 \end{pmatrix} & \big(\pi_{(\frac{1}{2},\frac{1}{2})}'(B_x)\big)^* &= \frac{i}{2}\begin{pmatrix} 0&1&0&0 \\ 1&0&0&0 \\ 0&0&0&1 \\ 0&0&1&0 \end{pmatrix} \\ \pi_{(\frac{1}{2},\frac{1}{2})}'(A_y) &= \frac{1}{2}\begin{pmatrix} 0&0&-1&0 \\ 0&0&0&-1 \\ 1&0&0&0 \\ 0&1&0&0 \end{pmatrix} & \big(\pi_{(\frac{1}{2},\frac{1}{2})}'(B_y)\big)^*&= \frac{1}{2}\begin{pmatrix} 0&-1&0&0 \\ 1&0&0&0 \\ 0&0&0&-1 \\ 0&0&1&0 \end{pmatrix} \\ \pi_{(\frac{1}{2},\frac{1}{2})}'(A_z) &= -\frac{i}{2}\begin{pmatrix} 1&0&0&0 \\ 0&1&0&0 \\ 0&0&-1&0 \\ 0&0&0&-1 \end{pmatrix} & \big(\pi_{(\frac{1}{2},\frac{1}{2})}'(B_z)\big)^* &= \frac{i}{2}\begin{pmatrix} 1&0&0&0 \\ 0&-1&0&0 \\ 0&0&1&0 \\ 0&0&0&-1 \end{pmatrix} \end{align*} We can then write out the matrices of the rotations and boosts $J_i$ and $K_i$ using $$ J_i = A_i + B_i \hspace{2cm} K_i = i(A_i - B_i). $$ \begin{align*} \pi'_{(\frac{1}{2},\frac{1}{2})}(J_x) &= \frac{i}{2}\begin{pmatrix} 0&1&-1&0 \\ 1&0&0&-1 \\ -1&0&0&1 \\ 0&-1&1&0 \end{pmatrix} & \pi'_{(\frac{1}{2},\frac{1}{2})}(K_x) &= \frac{1}{2}\begin{pmatrix} 0&1&1&0 \\ 1&0&0&1 \\ 1&0&0&1 \\ 0&1&1&0 \end{pmatrix} \\ \pi'_{(\frac{1}{2},\frac{1}{2})}(J_y) &= \frac{1}{2}\begin{pmatrix} 0&-1&-1&0 \\ 1&0&0&-1 \\ 1&0&0&-1 \\ 0&1&1&0 \end{pmatrix} & \pi'_{(\frac{1}{2},\frac{1}{2})}(K_y) &= \frac{i}{2}\begin{pmatrix} 0&1&-1&0 \\ -1&0&0&-1 \\ 1&0&0&1 \\ 0&1&-1&0 \end{pmatrix} \\ \pi'_{(\frac{1}{2},\frac{1}{2})}(J_z) &= \begin{pmatrix} 0&0&0&0 \\ 0&-i&0&0 \\ 0&0&i&0 \\ 0&0&0&0 \end{pmatrix} & \pi'_{(\frac{1}{2},\frac{1}{2})}(K_z) &= \begin{pmatrix} 1&0&0&0 \\ 0&0&0&0 \\ 0&0&0&0 \\ 0&0&0&-1 \end{pmatrix} \\ \end{align*} These are strange matrices, although we can make them look much more suggestive in another basis. Define the matrix \begin{equation*} U = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & -i & 0\\ 0 & 1 & i & 0 \\ 1 & 0 & 0 &-1 \end{pmatrix}. \end{equation*} Amazingly, \begin{equation*} U^{-1} \big( \pi'_{(\frac{1}{2},\frac{1}{2})}(L_i) \big) U = L_i \hspace{1cm}U^{-1} \big( \pi'_{(\frac{1}{2},\frac{1}{2})}(K_i) \big) U = K_i. \end{equation*} Therefore, the $(\tfrac{1}{2}, \tfrac{1}{2})$ representation is equivalent to the regular "vector" representation of $SO^+(1,3)$. However, these "vectors" live in $\mathbb{C}^4$, not $\mathbb{R}^4$, which people usually don't mention.