Hint:
$1$. Write down the power series for $\ln(1+x)$.
$2$. Substitute $-x$ for $x$ to find the power series for $\ln(1-x)$.
If you don't know the power series for $\ln(1+x)$, but you probably do, use the power series for $\frac{1}{1+t}$, and the fact that $\ln(1+x)=\int_0^x\frac{dt}{1+t}$. Expand, and integrate term by term.
$3$. We have $\ln\left(\frac{1+x}{1-x}\right)=\ln(1+x)-\ln(1-x)$.
The absolute value signs surrounding $\frac{1+x}{1-x}$ that you asked about are pointless. Since we are only working with $|x|<1$, the quantity $\frac{1+x}{1-x}$ is positive, so the absolute value signs do nothing.
Some detail: It is standard that $\frac{1}{1-u}=1+u+u^2+u^3+\cdots$, with the series converging when $|u|<1$. Put $u=-t$. We obtain $\frac{1}{1+t}=1-t+t^2-t^3+\cdots=\sum_{k=0}^\infty (-1)^kt^k.$ It follows that (for $|x|<1$) $\ln(1+x)=\int_0^x\frac{dt}{1+t}=\int_0^x\left(\sum_{k=0}^\infty (-1)^kt^k \right)dt=\sum_{k=0}^\infty(-1)^k\frac{x^{k+1}}{k+1}.$ We have now carried out Hint $1$.
Now substitute $-x$ for $x$ in our expression for $\ln(1+x)$. We get $\ln(1-x)=\sum_{k=0}^\infty(-1)^k\frac{(-x)^{k+1}}{k+1}=-\sum_{k=0}^\infty \frac{x^{k+1}}{k+1}.$ (The terms $(-1)^k$ and $(-1)^{k+1}$ have product $(-1)^{2k+1}$, which is identically equal to $-1$.) We have now carried out Hint $2$.
Now use Hint $3$. We get $\ln\left(\frac{1+x}{1-x}\right)=\sum_{k=0}^\infty \left[(-1)^k-(-1)\right]\frac{x^{k+1}}{k+1}.$ Consider the coefficient $[(-1)^{k}-(-1)]$ above. This is $2$ if $k$ is even and $0$ if $k$ is odd. To put it another way, we can put $k=2n$, since odd $k$ make no contribution to the sum. Thus $\ln\left(\frac{1+x}{1-x}\right)=\sum_{n=0}^\infty 2\frac{x^{2n+1}}{2n+1}.$