How can we better understand multiplicative inverse modulo something? How can we intuitively understand modulo multiplicative inverse?

Suppose we have an ring $\mathbb{Z}_{13}= \{0, 1, 2, 3, \ldots, 12\}$.

Each element except zero has corresponding multiplicative inverse.

Below is the mapping of inverse.

1 $\rightarrow$ 1

2 $\rightarrow$ 7

3 $\rightarrow$ 9

4 $\rightarrow$ 10

5 $\rightarrow$ 8

6 $\rightarrow$ 11

7 $\rightarrow$ 2

8 $\rightarrow$ 5

9 $\rightarrow$ 3

10 $\rightarrow$ 4

11 $\rightarrow$ 6 

12 $\rightarrow$ 12

Now, I want to consider division by 2, which here means that multiplication by 7.
However, some integer multiplied 7 becomes acutally the same result as division by 2 over real number field.
For example, the following holds.

4 * 7 = 28 = 2 mod 13;
6 * 7 = 42 = 3 mod 13;
etc.

On the other hand, other values do not produce the same result over real number fields. 

For example,
5 * 7 = 35 = 9 mod 13 (I want this value to be 2 or 3 since 5/2 = 2.5)

7 * 7 = 49 = 10 mod 13 (I want this value to be 3 or 4 since 7/2 = 3.5).

Why some values produce the same result as over the real number field, and the others do not??
Operations mod $n$ aren't guaranteed to preserve the order inherited from the reals.

So the fact that in $\mathbb{R}$, we have
$$2 &lt; {\small{\frac{5}{2}}} &lt; 3$$
doesn't imply
$$2 &lt; ({\small{\frac{5}{2}}}\;\text{mod}\;13) &lt; 3$$
However what is true is that, working mod $13$, we have
$$
{\small{\frac{5}{2}}} = 2+{\small{\frac{1}{2}}} = 2+7 = 9
\qquad\;\;\;
$$
and also
$${\small{\frac{5}{2}}} = 3-{\small{\frac{1}{2}}} = 3-7 = -4 = 9$$
What you seek is generally impossible. Suppose that $\,b\,$ is invertible $\!\bmod n,\,$ i.e. $\,\gcd(b,n)=1\,$ and let's calculate the fraction $\,a/b = ab^{-1}\pmod{\! n},\,$ where $\, 0 &lt; a,b &lt; n$.

By division $\,a = q\,b+r\ $ thus $\bmod n\!:\,\ a/b\equiv ab^{-1} \equiv (qb+r)b^{-1}\equiv q + \color{#c00}{rb^{-1}}$

When $\,r = 0\,$ we get $\!\bmod n\!:\ (qb)/b\equiv q,\,$ same as in $\Bbb R,\,$ when $\,a/b\in\Bbb Z\,$ is an exact quotient.

Else $\,\color{#0a0}{0&lt;r&lt;b}\,$ and you want $\,\color{#c00}{rb^{-1}}\equiv 0\,\ {\rm  or}\,\ 1\, \Longrightarrow\, \color{#0a0}{r\equiv  0\,\ {\rm or}\,\ b},\,$  contradiction.

Note $ $ It can be true for fractions with $\,a,b &gt; n\,$ if they reduce to exact quotients $(r= 0),\,$ e.g. 

$$\bmod 13\!:\ \ \dfrac{13}{27}\equiv \dfrac{0}1 = \left\lfloor \dfrac{13}{27}\right\rfloor,\ \ \dfrac{14}{27}\equiv \dfrac{1}1 = \left\lceil \dfrac{14}{27} \right\rceil\qquad\qquad\qquad $$