Let $y=f(x)=\arccos(1-x)$. Then $$1-x=\cos y=1-\frac{y^2}2+\frac{y^4}{24}+O(y^6),$$ so $$x=\frac{y^2}2-\frac{y^4}{24}+O(y^6).$$ Now clearly there is no actual Taylor series for $y$ about $x=0$ because $f'(0)$ does not exist. However, a generalized power series solution can be written down, known variously as the Frobenius method or the asymptotic expansion of $y=f(x)$ near $x=0$. Solving this equation formally:
$$2x=y^2\left(1-\frac{y^2}{12}+O(y^4)\right)\Rightarrow y=\sqrt{2x}\left(1-\frac{y^2}{12}+O(y^4)\right)^{-1/2}$$
Since $0<y\ll1$, $\frac{y^2}{12}\ll1$ so that we can use the binomial theorem $(1+x)^p=1+px+\cdots$ to get the next-leading order term:
$$y=\sqrt{2x}+\sqrt{2x}\frac{y^2}{24}+O(y^4)=\sqrt{2x}+\frac{\sqrt{2x}}{24}\left(\sqrt{2x}+\frac{\sqrt{2x}}{24}y^2+O(y^4)\right)^2+O(y^4)$$
$$=\sqrt{2x}+\frac{(2x)^{3/2}}{24}\left(1+\frac{1}{12}y^2+O(y^4)\right)+O(y^4)=\sqrt{2x}+\frac{(2x)^{3/2}}{24}+\frac{(2x)^{3/2}}{24\cdot 12}y^2+O(x^{3/2}y^4)+O(y^4)$$
Now, since $y=O(\sqrt x)$ (which follows from the leading order term), we can simplify all that to get $$y=\sqrt{2x}+\frac{(2x)^{3/2}}{24}+O(x^2).$$
I eventually managed to figure this out but for me the problem was not appreciating in full how stochastic calculus differs from ordinary calculus in this problem.
In ordinary calculus, the following is true
$df=\frac{\partial f}{\partial t}dt+\frac{\partial f}{\partial x}dx$
for a function $f$ that depends on $t$ and $x$. This fact can be derived by starting with the Taylor series for a function of 2 variables
$f(t,x)=f(a,b)+f_t(a,b)(t-a)+f_x(a,b)(x-b)...$ as I gave in my question.
Then consider
$f(t,x)-f(a,b)=f_t(a,b)(t-a)+f_x(a,b)(x-b)...$ as a and b approach t and x respectively. In the limit, all the second order and higher terms tend to zero much faster than the first order terms and are dropped. So in the limit, this becomes
$df(t,x)=f_t(t,x)dt+f_x(t,x)dx$
But with stochastic processes, the $dx$ actually contains something on the order of $\sqrt {dt}$ so then the $dx^2$ contains something on the order of $dt$, so then in the limit earlier, it contains a term which does not tend to zero any faster than $dx$ so it is not dropped.
Best Answer
We have for the first derivative $g^{(1)}(t)$
$$\begin{align} g^{(1)}(t)&=-\frac{2t}{\sqrt{1-(1-t^2)^2}}\\\\ &=-\frac{2}{\sqrt{2-t^2}}\tag 1 \end{align}$$
Differentiating the right-hand side of $(1)$, we obtain the second derivative, $g^{(2)}(t)$
$$\begin{align} g^{(2)}(t)&=-\frac{2t}{(2-t^2)^{3/2}}\tag 2 \end{align}$$
Continuing, we have for $g^{(3)}(t)$
$$\begin{align} g^{(3)}(t)&=-\frac{4(t^2+1)}{(2-t^2)^{5/2}}\tag 3 \end{align}$$
And finally, we have for $g^{(4)}(t)$
$$\begin{align} g^{(4)}(t)&=-\frac{12t(t^2+3)}{(2-t^2)^{7/2}}\tag 4 \end{align}$$
We evaluate $(1)-(4)$ at $t=0$ and form the expansion
$$\bbox[5px,border:2px solid #C0A000]{\arcsin(1-x)=\frac{\pi}{2}-\sqrt{2}x^{1/2}-\frac{\sqrt{2}}{12}x^{3/2}+O\left(x^{5/2}\right)}$$