How is it, that $\sqrt{x^2}$ is not $ x$, but $|x|$? As far as I see, $\sqrt{x^2}$ is not $x$, but $|x|$, meaning the "absolute". I totally get this, because $x^2$ is positive, if $x$ is negative, so $\sqrt{y}$, whether $y = 10^2$ or $y = -10^2$: $y$ is positive.

But then I remember that $\sqrt(x)$ is the same as $x^{1/2}$ and thus, $\sqrt{x^2}$ is the same as $x^{1}$. 

So, as far as I get it, $\sqrt{x^2} = x^{2/2}$.

But then, I can cancel $\frac{2}{2}$ to $1$. So: $\sqrt{x^2} = x^1 = x$.

Where does the absolute value in this derivation of the calculation come from?

I get it why its there: Because $(-x)^2 = x^2$, so the radicand is always positive. But from $\sqrt{x^n} = x^{n/2}$ I dont understand how it gets there.
This is why you need to pay attention to the hypotheses of a theorem! The exponent laws you've learned are almost surely only for integer exponents or for positive bases: until you have learned multi-valued functions and complex exponentiation, you should always be suspicious whenever you're both using non-integers in exponents and negative numbers in bases in the same problem. (presumably, you'd understand the nuances once you've fully learned how to deal with complex analysis)
$√x$ = $(x)^{1/2}$ represents only the principle value of square root of x. Its the exponential representation of principle square root.

Principle square root: The non-negative (positive) square root of a number is called its Principal Square Root.

Thus, $√(x^2)$ = $(x^2)^{1/2}$ = x is the principle square root (positive)

While, $√(x^2)$ = $±x$ ; both positive and negaive square roots.