Let $U_i:=\{p\in S^1:x_i\ne 0\}$, $i=1,2$, be two open sets of $S^1$. Define $$ \omega_p := \begin{cases} \Bigl(\bigl(-\frac{dx_2}{x_1}\bigr)|_{U_1}\Bigr)_p\,\,\,\;\text{if}\,\,p\in U_1\\ \Bigl(\bigl(\frac{dx_1}{x_2}\bigr)|_{U_2}\Bigr)_p\,\,\,\;\text{if}\,\,p\in U_2\\ \end{cases} $$ I want to prove that this is a well-defined $1$-form on $S^1$.
What I have to do is to show that for $p\in U_1\cap U_2$, $$ \Bigl(\bigl(-\frac{dx_2}{x_1}\bigr)|_{U_1}\Bigr)_p=\Bigl(\bigl(\frac{dx_1}{x_2}\bigr)|_{U_2}\Bigr)_p $$ right?
I can write $p=(\cos\theta,\sin\theta)$. Let $v\in T_p S^1$, then $v=-\sin\theta\frac{\partial}{\partial x_1}+\cos\theta\frac{\partial}{\partial x_2}$.
Let's apply $\omega$ to $v$, then $$ \Bigl(\bigl(-\frac{dx_2}{x_1}\bigr)|_{U_1}\Bigr)_p(v)=-\frac{\cos\theta}{\cos\theta}=-1\\ \Bigl(\bigl(\frac{dx_1}{x_2}\bigr)|_{U_2}\Bigr)_p= -\frac{\sin\theta}{\sin\theta}=-1 $$ Is it okay?
Yes, in spirit this is OK. You should be specifying, however, that $U_i = \{x\in S^1: x_i\ne 0\}$. And your "then $v=$ ..." is not totally correct, as $v$ could be any scalar multiple of this vector.
An alternative approach (more amenable to generalization) is to observe that since $S^1$ is defined by the equation $x_1^2+x_2^2=1$, the $1$-form $$\frac 12 d(x_1^2+x_2^2) = x_1\,dx_1+x_2\,dx_2$$ is identically $0$ on $S^1$. It follows by simple algebra that $\dfrac{dx_1}{x_2}+ \dfrac{dx_2}{x_1} = 0$ wherever $x_1x_2\ne 0$, i.e., on $U_1\cap U_2$. It's better to learn to work with differential forms without always having to evaluate them on tangent vectors.