$(\Rightarrow)$ If $M^m\times N^n$ is orientable, there is a volume form $\eta\in\Omega^{n+m}(M\times N)$. For any fixed point $q\in N$, take a basis $\{w_1,...,w_n\}$ for $T_qN$. Now define $\omega\in\Omega^m(M)$ by:
$$\omega(X_1,...,X_m):=\eta_{(\cdot,q)}\left(X_1,...,X_m,w_1,...,w_n\right)$$
We will prove that $\omega$ is a volume form. For a fixed $p$, take a basis $\{v_1,...,v_m\}$ for $T_pM$. Using the identification $T_{(p,q)}M\times N\equiv T_pM\oplus T_qN$, then $\{v_1,...,v_m,w_1,...,w_n\}$ is a basis for $T_{(p,q)}M\times N$. Since $\eta$ is a volume form, $\omega_p(v_1,...,v_m)=\eta_{(p,q)}(v_1,...,v_m,w_1,...,w_n)\neq 0$, which means $\omega$ is a volume form, so $M$ is orientable. By a similar argument, $N$ is orientable.$_\blacksquare$
$(\Leftarrow)$ If $M,N$ are orientable, there are volume forms $\omega\in\Omega^m(M), \sigma\in\Omega^n(N)$. For the natural projections $\pi_M:M\times N\to M$ and $\pi_N:M\times N\to N$, define:
$$\eta:=\pi_M^*(\omega)\wedge \pi_N^*(\sigma)\in\Omega^{n+m}(M\times N)$$
We will prove $\eta$ is a volume form. For a fixed $x=(p,q)\in M\times N$, take basis $\{v_1,...,v_n\}$ for $T_pM$ and $\{w_1,...,w_n\}$ for $T_qN$. Because $\omega$ and $\sigma$ are volume forms, we have $\omega_p(v_1,...,v_m)\neq 0$ and $\sigma_q(w_1,...,w_n)\neq 0$. Since $\{v_1,...,v_m,w_1,...,w_n\}$ is a basis for $T_x(M\times N)$, it's enough to check that $\eta_x(v_1,...,v_m,w_1,...,w_n)\neq 0$. Indeed, noticing that $(d\pi_M)_x(v_i)=v_i$, $(d\pi_M)_x(w_j)=0$, $(d\pi_N)_x(v_i)=0$ and $(d\pi_N)_x(w_j)=w_j$, we have:
$$\eta_x(v_1,...,v_m,w_1,...,w_n)=\underbrace{\omega_p(v_1,...,v_m)}_{\neq 0}\,\underbrace{\sigma_q(w_1,...,w_n)}_{\neq 0}\neq 0\,\,_\blacksquare$$
While it might be a good exercise, there's no need here to work in coordinate charts. If $F:\mathbb{R}^{2n}\to\mathbb{R}$ is $F(p) = \|p\|^2$, the sphere $S^{2n-1}$ is defined as $F^{-1}(1)$.
Define a vector field $V$ on $\mathbb{R}^{2n}$ by $$V_{(x,y)} = \sum_i -y^i\partial x_i + x^i\partial y_i$$
As the restriction of a smooth function to a smooth submanifold is again smooth, it will suffice to show that $X$ is the restriction to $S^{2n-1}$ of $V$. We will see this by showing that $X$ is perpendicular to the gradient of $F$ at a point $(x,y)\in S^{2n-1}$.
Note that $\nabla F_p = \sum 2p_i\partial_i$. Then:
$$\begin{align*}
\langle V_{(x,y)}, \nabla F_{(x,y)}\rangle &= \bigg\langle \sum_i -y_i\partial x_i + x_i \partial y_i , \sum_i 2x_i \partial x_i + 2y_i \partial y_i\bigg\rangle \\
&= \sum_i -2y_ix_i + 2x_iy_i \\
&= 0
\end{align*}$$
As $V$ is nonvanishing away from zero, $X$ is manifestly a nonvanishing smooth vector field on $S^{2n-1}$.
Best Answer
First, you need to figure out how to change between the two coordinate charts $x_1$ and $x_2$. Let $(y,z)$ be a point on the unit circle $y^2+z^2=1$. Using triangle ratios, you can find $x_1 = \dfrac y{1-z}$ and $x_2 = \dfrac y{1+z}$. Then $x_1x_2 = \dfrac{y^2}{1-z^2} = 1$, since $y^2+z^2=1$. Thus, whenever both $x_1,x_2$ are both defined, $x_1x_2 = 1$.
Now we want to get $\dfrac\partial{\partial x_1}$ in terms of $x_2$. By the chain rule, this is $\dfrac{\partial x_1}{\partial x_2}\dfrac\partial{\partial x_2}$. Then using $x_1 = \dfrac1{x_2}$, $\dfrac{\partial x_1}{\partial x_2} = \dfrac{-1}{x_2^2}$. Thus $\dfrac\partial{\partial x_1} = \dfrac{-1}{x_2^2}\dfrac\partial{\partial x_2}$. The point $N$ corresponds to $x_2 = 0$, so we see that this quantity is diverging. So no, the vector field is not smooth.