Why do we invert then multiply when dividing fractions?

Several answers focused on the fact that division is really just multiplication with the inverse, and that is perfectly fine.

Here's another aproach:

When you calculate what $\frac{a}{b}$ is, what are you really doing? You are asking yourself

"How many pieces do I get if I divide $a$ into $b$ equal pieces?"

so you are asking yourself

"by how much must I multiply $b$ to get $a$?"

For example, if you are calculating $\frac{10}2$, you want to know how many groups of $2$ you can make out of $10$ people, or, equivalently, what you need to multiply by $2$ to get $10$, and the answer is $5$ because $2\cdot 5=10$.

So, the bottom line here is that $x=\frac{a}{b}$ is defined as the unique number that satisfies the equation $bx = a$.

Using this definition, you can now see what happens if $b\frac12$. Well, in that case, by how much must I multiply $\frac 12$ to get $a$? In other words, what is the solution to the equation $$\frac12 \cdot x = a?$$

You can easily see that the answer is obvious when you multiply the equation by $2$ to get $x=2a$.

$\frac{x}{\frac{1}{2}}=\frac{x}{\frac{1}{2}}\cdot1=\frac{x}{\frac{1}{2}}\cdot\frac{2}{2}=\frac{x\cdot 2}{\frac{1}{2}\cdot2}=\frac{2x}{1}$

Generally,

$\frac{a}{\frac{b}{c}}=\frac{a}{\frac{b}{c}}\cdot\frac{c}{c}=\frac{ac}{b}=a\cdot\frac{c}{b}$

Unfortunately, nobody tells you the definition of $x/y$ nowadays. (Did they ever?) The simplest definition is surely that if $x$ is a real number and $y$ is a non-zero real number, then $x/y$ is--by definition-- a mere shorthand for the expression $x y^{-1}.$ So whenever $y$ is non-zero, we have:

$$\frac{x}{y} = x/y = xy^{-1}$$

This viewpoint is very powerful; for example, your question now becomes very easy to answer. Observe that if $a$ is a real number and $c$ and $d$ are non-zero real numbers, then we may argue as follows:

$$\frac{a}{c/d} = \frac{a}{c d^{-1}} = a(cd^{-1})^{-1} = a(c^{-1}d) = adc^{-1} = \frac{ad}{c}$$

The above chain of reasoning makes use of the following rules about raising to the power of $-1$, which hold for all real numbers distinct from $0$:

1. $(xy)^{-1} = x^{-1} y^{-1}$
2. $(x^{-1})^{-1} = x$

Other important laws (that I didn't use) include:

3. $x x^{-1} = 1$
4. $1x = x$ (this last one holds even if $x$ equals $0$.)

In summary:

Theorem. For all real numbers $a$ and all non-zero real numbers $c$ and $d$, we have: $$\frac{a}{c/d} = \frac{ad}{c}$$

Applying this to your problem:

$$\frac{x}{1/2} = \frac{x \cdot 2}{1} = 2x$$

The take home message is that before trying to understand division, you should try to understand raising to the power of $-1$. The expressions $x/y$ and $$\frac{x}{y}$$ are best viewed as nifty notation for $x y^{-1}$, nothing more.