Multiplicative Inverse?

There are two problems, one each depending on what $\frac{1}{2}$ means.

1. We can define $\frac{1}{a}$ to be "the [unique] number such that $a\times\frac{1}{a}=1$, if it exists" in whatever system we are working on (that is, "number" here would mean "number modulo $N$"). But then you cannot assume that such a thing as $\frac{1}{2}\pmod{N}$ exists in the first place. You must prove it exists.

   Note that they don't always exist: for example, $2$ has no multiplicative inverse in the integers either.

   In this situation, it is true that $\frac{1}{2}\pmod{6}$ is a multiplicative inverse of $2\pmod{6}$, if it exists. But in fact, no such thing exists. Just like an even prime number greater than $2$ would be congruent to $0$ modulo $2$ if it existed, but no such thing exists.

2. If by $\frac{1}{2}$ you mean the rational number $\frac{1}{2}$, then $\frac{1}{2}\pmod{6}$ makes no sense in the integers modulo $6$: we only allow integers! That is, when we write things like $a\pmod{N}$, we are implicitly asserting that $a$ is an integer. We cannot do that with $\frac{1}{2}$.

   To see that there cannot exist an integer $x$ such that $2x\equiv 1\pmod{6}$, note that $2x-1$ is always odd, so it is never a multiple of $6$; hence, $2x$ can never be congruent to $1$ modulo $6$, no matter what integer $x$ is.

One thing you're missing is that you shouldn't be referring to such alleged objects as $2\bmod 6$, etc. $$ \text{Right:}\quad (x \equiv y)\pmod n $$ $$ \text{Wrong:}\quad x \equiv \Big( y \bmod n \Big) $$

To say that $x$ and $y$ are mod-$n$ congruent to each other means their difference is a multiple of $n$. Thus for example, $(69\equiv 62) \pmod 7$.

Now observe that $(3\cdot5 \equiv 1)\pmod 7$, so $3$ and $5$ are each other's mod-$7$ reciprocals.