Denote by $L$ the Lorentz group $O(1,3)$ and by $L_+=\{M\in L:\det M=1,M_{00}\ge 1\}$. This is the connected component of the identity in $L$, but we don't care in this exercise.
First, we define a map $$\widehat{\cdot}:\mathbb{R}^4\to H=\{M\in\mathbb{C}^{2\times 2}:M=M^{\dagger}\}, \widehat{(x_0,x_1,x_2,x_3)}=\begin{pmatrix}x_0+x_3&x_1-ix_2\\x_1+ix_2&x_0-x_3\end{pmatrix}.$$ This is a bijection.
We define a map $f:\text{SL}(2,\mathbb{C})\to L_+$ as $f(M):\mathbb{R}^4\to\mathbb{R}^4, x\mapsto x'$, where $x'$ is unique such that $\widehat{x'}=M\widehat{x}M^{\dagger}$.
I have shown that $f$ is a group homomorphism. But why is $f$ well-defined, i.e. has image in $L_+$, i.e. why is $f(M)\in L$, $\det f(M)=1$ and $f(M)_{00}\ge 1$?
The main thing to observe is that $SL(2,C)$ acts on the (real 4-dimensional) vector space $V$ of hermitian forms (in two variables) by change of coordinates. If a form $q$ has Gram matrix $M$ then the action of $A^{-1}\in SL(2,C)$ on $q$ is described by $$ M\mapsto A^* M A. $$ (You are using the reverse of this formula for some reason.) This action preserves the determinant function $M\mapsto det(M)$ (since $det(A)=1$) which, as you probably know, is quadratic. hence, you get a representation $\rho: SL(2,C)\to O(V,q)$: The space of automorphisms of $V$ preserving the form $q$. Now, if you compute the signature of the quadratic form $det$ on $V$ you get that it is $(1,3)$. Hence, you got a representation $SL(2,C)\to O(1,3)$. Since $SL(2,C)$ is connected and $\rho$ is continuous, the image lands in the identity component of $O(1,3)$. If you need more details, let me know.