I refer to pp. 89-90 of "Partial Differential Equations of Mathematical Physics and Integral Equations" by Ronald B. Guenter & John W. Lee
Given $u_{tt}=c^2u_{xx}$ for $x\in \mathbb{R}$ and $t>0$
$u(x,0)=f(x)$ and $u_t(x,0)=g(x)$ for $x\in \mathbb{R}$
The characteristic curves are $x+ct=$constant and $x-ct=$constant, so we choose the transformation $\alpha=x+ct$ and $\beta=x-ct$ to obtain $u_{\alpha\beta}=0$, then integrating, we get $u=A(x+ct)+B(x-ct)$ for arbitrary single valued $A,B$. In the end, we get $A(x)=(1/2)f(x)+(1/2c)\int^x_0g(s)ds+D/2$ and $B(x)=(1/2)f(x)-(1/2c)\int^x_0g(s)ds-D/2$.
My question is, what if I chose $\alpha=x+ct$ and $\beta=-x+ct$? Then $u=A(\alpha(x,t))+B(\beta(x,t))$,
so $u(x,0)=A(x)+B(-x)$.
$u_t(x,t)=A_\alpha(\alpha(x,t))\alpha_t(x,t)+B_\beta(\beta(x,t))\beta_t(x,t)=cA_\alpha(\alpha(x,t))+cB_\beta(\beta(x,t))$
so $g(x)=u_t(x,0)=cA_\alpha(\alpha(x,0))+cB_\beta(\beta(x,0))=cA_\alpha(x)+cB_\beta(-x)$
We also have $\dfrac{d}{d\alpha}=\dfrac{\partial}{\partial x}\dfrac{\partial x}{\partial \alpha}$ and $\dfrac{d}{d\beta}=\dfrac{\partial}{\partial x}\dfrac{\partial x}{\partial \beta}$ and $\dfrac{\partial \beta}{\partial x}=-1$ and $\dfrac{\partial \alpha}{\partial x}=1$
Hence, $g(x)=cA_\alpha(x)+cB_\beta(-x)=cA_x(x)-cB_x(-x)$ Integrating the last equation, $\int_0^xg(s)ds=cA(x)+cB(-x)$.
So $A(x)+B(-x)=(1/c)\int_0^xg(s)ds=f(x)$, in which case I cannot solve for $A(x)$ and $B(x)$ in terms of $f$ and $g$. I thought that characteristic curves of a PDE is not unique, but in this calculation, choice of one characteristic curve, namely $\beta=x-ct$ works, whereas another choice, namely $\beta=-x+ct$ does not work, where am I mistaken?
Problems with your approach
Firstly, $$ \frac{d}{d\alpha} = \frac{\partial x} {\partial \alpha} \frac{\partial } {\partial x} + \frac{\partial t} {\partial \alpha} \frac{\partial } {\partial t}, $$ you missed the second term. Also $$ \frac{\partial x} {\partial \alpha} \neq \left(\frac {\partial \alpha} {\partial x}\right) ^{-1}, $$ the former is evaluated at constant $\beta$, the latter at constant $t$. The correct way is to note that $$ x=\frac{\alpha-\beta} {2} $$ thus $$ \frac{\partial x} {\partial \alpha} =\frac{1} {2}. $$ If you attempt to continue this approach you'll just get a massive mess.
A better approach
You have that $$ g(x) = cA'(x) + c B'(-x) $$ which can be integrated to yield $$ \int_0^x g(s) ds = [cA(s) - cB(-s)]_0^x. $$ From here you should easily arrive at your previous solution (up to a sign).