I read that the modulo operation finds remainder after division.
My misconseption here is about the remainder.
Is the remainder the last digit of the result? I thought it was the first.

Example: $5/7 = 0.7142857142857142857...$

$5 \mod 8 = 7$.

$7/3 = 2.33333..$ but $7 \mod 3 = 1$. Remainder of the first result is point 3333...

$9/4 = 2.25$ $9 \mod 4 = 1$. Ok same as first but different remainder.

$9/7 = 1.2857142857142...$ $9 \mod 7 = 2$. What? Remainder is two, the first digit. This is probably "random" for reals? But not for integers?

$111111111$ : If there are nine ones we should divide by $7$, it gives $2$ remnants. Since $9-2=7$ ones.
So using modulo operation on the reals is not something one should do?
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When you want to compute $a \bmod b$ with $a,b$ positive integers, you do division like you learned in school and write $a=qb+r$ where $q$ is the quotient and $r$ is the remainder. We typically choose $r$ to be in the range $0$ to $b-1$. Then we write $r=a \bmod b$, so is you want $1234 \bmod 7$, you write $1234=176\cdot 7 + 2$, so $1234 \bmod 7=2$. You don't need to do anything with the decimal expansion. 

The main time I have seen the modulo operation used with the modulus $b$ other than an integer is reducing angles of a circle when the angles are measured in radians. You compute an angle, then take it $\mod 2\pi$ to put the angle in the range $[0,2\pi)$. You can use other moduli, but this is the common one.
Let's examine a couple of your examples.

First $\dfrac{7}3 = \color{#0a0}2.333\ldots = \color{#0a0}2 + \dfrac{\color{#c00}1}3\ $ has quotient $\color{#0a0}2$ and remainder $\color{#c00}1$

without fractions: $\ 7 = \color{#0a0}2\cdot 3 + \color{#c00}1$

and $\,\ \dfrac{9}7 = \color{#0a0}1.286\ldots = \color{#0a0}1 + \dfrac{\color{#c00}2}7\ $ has quotient $\color{#0a0}1$ and remainder $\color{#c00}2$

without fractions: $\ 9 = \color{#0a0}1\cdot 7 + \color{#c00}2$

Generally we have $\, a = \color{#0a0}q\cdot b + \color{#c00}r\ $ with remainder $\ \color{#c00}r = a\bmod b\,$ obeying $\, 0\le r < b$

which is obtained by (long) dividing $a$ by $b\,$ (Euclidean division with remainder)

This remainder $r$ is the *least* natural of the form $\, a-q\,b\,$ for some integer $q$.

Said procedurally: continually add/subtract $b$ from $a$ till you land in the interval $[0,b)$ 

Note that if $a, b\neq 0$ are reals then the prior procedure still works fine.