We know the power series expansion for $\dfrac{1}{1+s}$. The power series for $\log(1+t)$ can then be obtained efficiently by term by term integration. But the way that was used in the post was not much harder.
Then it is easy to find the power series expansion of $f(x)$ by division and term by term integration.
There are easier and better ways to approach the error term. The Lagrange form of the remainder often gives thoroughly bad results, because it involves an unknown $\xi$ that one may end up having to be excessively pessimistic about.
For positive $x$, note that since for convergence we need $x \le 1$, our series for $f(x)$ is an alternating series. Thus the error made by truncating anywhere has absolute value less than the first "neglected" term. That quickly gives us an error estimate that is more useful than the Lagrange expression. For negative $x$, we don't have the nice alternating series feature, but a good error estimate can be found by looking at the tail and comparing it with a geometric series.
Added detail about error term: We get that $f(x)$ has the power series expansion
$$x-\frac{x^2}{4}+\frac{x^3}{9}-\frac{x^4}{16}+ \cdots +(-1)^{n+1}\frac{x^n}{n^2}+\cdots.$$
By standard convergence tests, this diverges if $|x|>1$. There is somewhat reluctant convergence at $x=\pm 1$, and brisker convergence if $|x|<1$.
If $0<x\le 1$, we have an alternating series (the terms alternate in sign, and decrease steadily in absolute value, with limit $0$). For alternating series, the error made by keeping everything up to the $m$-th term and throwing away the rest is of the same sign as the first "neglected" term, and has absolute value less than the absolute value of the first neglected term.
For example, if we keep everything up to the term $-x^6/36$, and throw away the rest, then our estimate for $f(x)$ is an underestimate and the error is less than $x^7/49$.
The situation is different if we use our series to evaluate $f(x)$ for negative $x$. For then the series is no longer an alternating series. Suppose $x$ is negative (but $>-1$). Suppose also that, for example, we throw away all the terms from $x^7/49$ on. To get rid of minus signs, let $w=-x$. Then our error has absolute value
$$\frac{w^7}{49}+\frac{w^8}{64}+\frac{w^9}{81}+\cdots.$$
This is less than
$$\frac{w^7}{49}+\frac{w^8}{49}+\frac{w^9}{49}+\cdots.$$
Take the common factor $\frac{w^7}{49}$ out. What's left is the good old $1+w+w^2+\cdots$. So the error has absolute value less than
$$\frac{w^7}{49}\cdot \frac{1}{1-w}.$$
This turns out to be a pretty good estimate. The performance of Lagrange Remainder estimates is very bad in this kind of situation.
Neither is more correct. You can ask for the Taylor series about any point. It might or might not exist. If it exists, it will converge within a radius of the distance to the nearest singularity. Both of your functions are real analytic, meaning the Taylor series at both points exists and equals the function, everywhere.
The formula for the Taylor series centered at the point $a$, is $f(x)=\sum_{n\ge0}f^{(n)}(a)(x-a)^n/n!$.
To take a different example, the inverse of $e^x$, $\ln x$, it winds up making a bit more of a difference where you ask for the Taylor series, since it isn't always defined. One often considers $\ln(1+x)$, to avoid the singularity at $0$.
Best Answer
Airy functions are analytic on the complex plane. You can form their Taylor series about $x=a$ just as you would any other function. So for $x\approx a$, $$ \text{Ai}(x)=\text{Ai}(a)+ \text{Ai}'(a)(x-a)+\frac{1}{2}\text{Ai}''(a)(x-a)^2 +\frac{1}{6} \text{Ai}'''(a)(x-a)^3+\cdots $$
For example, the Taylor series for $\text{Ai}(x)$ about $x=0$ is $$ \text{Ai}(x)={1\over 3^{2/3}\pi}\sum_{k=0}^\infty {1\over k!}\Gamma\left({k+1\over 3}\right)\sin\left({2\pi (k+1)\over 3}\right)\left(\root 3 \of 3\,z\right)^k. $$