You may think about $(x_1,\ldots,x_n)$ as a realization of $n$ independent copies of $X$. Basically, there is a probability space $(\Omega,\mathcal{F},\mathsf{P})$ in the background so that $(x_1,\ldots,x_n)=(X_1(\omega)\ldots,X_n(\omega))$ for some $\omega\in\Omega$, which is chosen randomly according to $\mathsf{P}$. Then the statement "independent and identically-distributed realizations of the same random variable" doesn't make sense. Although, sometimes $(x_1,\ldots,x_n)$ is referred to as a random sample from a particular distribution (e.g. $F_X$).
The answer is
$$
n!\prod_{i=1}^n g(x_i,y_i)\mathbf{1}(x_1<x_2<\ldots<x_n).
$$
To see this, first observe that
$$
\mathbb{P}(X_{(i)}\leq x_i, Y_{(i)}\leq y_i, 1 \leq i \leq n)=n!\mathbb{P}(X_1<X_2<\ldots<X_n \text{ and } X_i \leq x_i, Y_i \leq y_i, 1 \leq i \leq n).
$$
The term on the right hand side can be written as
$$
\int_{-\infty}^{x_n}\int_{-\infty}^{y_n} \left\lbrace \ldots \left[
\int_{-\infty}^{\tilde{x}_3 \wedge x_2}\int_{-\infty}^{y_2}G(x_1 \wedge \tilde{x}_2,v_1)g(\tilde{x}_2,\tilde{y}_2)d\tilde{x}_2d\tilde{y}_2
\right] \ldots\right\rbrace g(\tilde{x}_n, \tilde{y}_n)d\tilde{x}_n d \tilde{y}_n\\
=\int_{-\infty}^{x_n}\int_{-\infty}^{y_n} \left\lbrace \ldots \left[
\int_{x_1}^{\tilde{x}_3 \wedge x_2}\int_{-\infty}^{y_2}G(x_1,v_1)g(\tilde{x}_2,\tilde{y}_2)d\tilde{x}_2d\tilde{y}_2
\right] \ldots\right\rbrace g(\tilde{x}_n, \tilde{y}_n)d\tilde{x}_n d \tilde{y}_n
+R_k(\mathbf{x}_{-2}, \mathbf{y})
$$
where $R_k(\mathbf{x}_{-2}, \mathbf{y})$ is a reminder term depending only on $\mathbf{x}_{-2}=(x_1, x_3, \ldots,x_n)$ and $\mathbf{y}=(y_1, \ldots,y_n)$. The term on the left-hand side can be further re-expressed as
$$
\int_{-\infty}^{x_n}\int_{-\infty}^{y_n} \left\lbrace \ldots \left[
\int_{x_2}^{\tilde{x}_4 \wedge x_3}\int_{-\infty}^{y_3}G(x_1,v_1)G(x_2,v_2)g(\tilde{x}_3,\tilde{y}_3)d\tilde{x}_3d\tilde{y}_3
\right] \ldots\right\rbrace g(\tilde{x}_n, \tilde{y}_n)d\tilde{x}_n d \tilde{y}_n
+R_k'(\mathbf{x}_{-2}, \mathbf{y})+R_k''(\mathbf{x}_{-3}, \mathbf{y})
$$
where the reminder terms $R_k'(\mathbf{x}_{-2}, \mathbf{y})$ and $R_k''(\mathbf{x}_{-3}, \mathbf{y})$ do not depend on $x_2$ and $x_3$, respectively. Iterating the procedure, we finally obtain that
$$
n!\mathbb{P}(X_1<X_2<\ldots<X_n \text{ and } X_i \leq x_i, Y_i \leq y_i, 1 \leq i \leq n)=n! \prod_{i=1}^nG(x_i,y_i) + R_n'''(\mathbf{x}_{-n},\mathbf{y})
$$
where the reminder term $ R_n'''(\mathbf{x}_{-n},\mathbf{y})$ does not depend on $x_n$ and accounts for all the reminder terms iteratively produced. Therefore, differentiating with respect to $x_1, \ldots, x_n, y_1, \ldots,y_n$, we are left with the expression in the first display.
Best Answer
As I see, you have a problem with just the definition.
Note that $X_1,...,X_n:\Omega \to \mathbb R$ are random variables (that is measurable functions).
For given $\omega \in \Omega$, you have certain values $X_1(\omega),...,X_n(\omega)$.
Those certain values can be sorted (maybe sorting is not unique). In other words, there exists permutation $\pi_{\omega}:\{1,...,n\} \to \{1,...,n\}$ such that:
$$ X_{\pi_{\omega}(1)}(\omega) \le ... \le X_{\pi_{\omega}(n)}(\omega)$$
Now, the point is, you define $X_{(k)}:\Omega \to \mathbb R$ with the formula: $$ X_{(k)}(\omega) = X_{\pi_{\omega}(k)}(\omega)$$