The general setting for defining a cross product of two vectors is as follows. Consider an $3$-dimensional, oriented, pseudo inner-product space $(V,\text{Or}, g)$, and let $\mu$ be the volume form on $V$ defined with respect to the given orientation and $g$ (see this question and answer of mine for more about the definition of this volume form).
Now, given $v,w\in V$, we define
\begin{align}
v\times w&:= g^{\sharp}\left(\mu(v,w,\cdot)\right)
\end{align}
Here, $g^{\flat}:V\to V^*$, $x\mapsto g(x,\cdot)$ is the flat-mapping, which one can easily show is an isomorphism, and $g^{\sharp}:V^*\to V$ is the inverse isomorphism. Just so we're clear: $\mu$ is the volume form on $V$, which means it is an alternating multilinear mapping $V\times V\times V\to\Bbb{R}$, so by feeding in $v,w$, we have $\mu(v,w,\cdot)\in V^*$, hence by applying $g^{\sharp}$ to it, we get an element of $V$. (and in the case of $\Bbb{R}^3$ with the standard inner product, we have $\mu=\det$ being the usual determinant of a matrix thought of as an alternating $(0,3)$-tensor on $\Bbb{R}^3$... and it coincides with the usual definition).
Now, extracting the components is basic linear algebra. Let $\{f_1,f_2,f_3\}$ be a positively-oriented (otherwise, there will be an extra minus sign at the end) basis (we make no assumptions about orthogonality) of $V$, and let $\{\phi^1,\phi^2,\phi^3\}$ be the dual basis for $V^*$. Then, as shown in my linked answer, one has $\mu= \sqrt{|g|_F}\,\phi^1\wedge \phi^2\wedge \phi^3$, where the square root is short for the following $3\times 3$ determinant:
\begin{align}
\sqrt{|g|_F}:= \sqrt{|\det [g(f_i,f_j)]_{i,j=1}^3|}
\end{align}
Also, I shall use the notation $g_{ab}=g(f_a,f_b)$ and $g^{ab}$ shall denote the $(a,b)$-entry of the matrix inverse to $[g_{ij}]_{i,j=1}^3$.
I hope you remember how elements of $V$ and $V^*$ can be expanded in terms of their bases. In what follows, all indices range over $1,2,3$, regardless of whether I use Latin/Greek indices. We have
\begin{align}
v\times w&:=g^{\sharp}\bigg(\mu(v,w,\cdot)\bigg)\\
&=g^{\sharp}\bigg(\mu(v,w,f_{\alpha})\phi^{\alpha}\bigg)\\
&=\mu(v,w,f_{\alpha})\cdot g^{\sharp}(\phi^{\alpha})\tag{$g^{\sharp}$ is linear}\\
&= \mu(v^{\beta}f_{\beta},w^{\gamma}f_{\gamma},f_{\alpha})\cdot g^{i\alpha}f_i\\
&= \mu(f_{\beta},f_{\gamma},f_{\alpha})\cdot g^{i\alpha}v^{\beta}w^{\gamma}\,f_i\\
&=\left(\sqrt{|g|_F}\,\phi^1\wedge \phi^2\wedge \phi^3\right)(f_{\beta},f_{\gamma},f_{\alpha})\cdot g^{i\alpha}v^{\beta}w^{\gamma}\,f_i\\
&= \sqrt{|g|_F}\,\, \varepsilon_{\beta\gamma\alpha}\cdot g^{i\alpha}v^{\beta}w^{\gamma}\,f_i
\end{align}
In the last line, I used the fact that $\phi^1\wedge\phi^2\wedge\phi^3(f_{\beta},f_{\gamma},f_{\alpha})=\det(e_{\beta},e_{\gamma},e_{\alpha})=\varepsilon_{\beta\gamma\alpha}$, where $e_1,e_2,e_3$ are the standard basis vectors of $\Bbb{R}^3$ (the reason why this holds, you just unwind the definition of the wedge product, and use the fact that the $\phi$'s are dual to the $f$'s, so eventually, determinants start popping up).
In other words, if we take a look at the $i^{th}$ component only, we have
\begin{align}
(v\times w)^i&=\sqrt{|g|_F}\,\, \varepsilon_{\beta\gamma\alpha}\cdot g^{i\alpha}v^{\beta}w^{\gamma}\\
&=\sqrt{|g|_F}\,\,g^{i\alpha}\varepsilon_{\alpha\beta\gamma}v^{\beta}w^{\gamma},
\end{align}
where in the last line, I permuted the indices in the Levi-Civita symbol appropriately to make things "look nice". Note that if $g$ is a proper inner-product and the vectors $\{f_1,f_2,f_3\}$ are orthonormal, then the determinant equals $1$ and we have a Kronecker delta, so
\begin{align}
(v\times w)^i&=\sum_{\beta,\gamma=1}^3\varepsilon_{i\beta\gamma}v^{\beta}w^{\gamma},
\end{align}
which is exactly what we expect
