<s>[INST] Different methods to multiply The sum and the multiplication in $\mathbb{Z}_p$ correspond to the the sum and the multiplication of powerseries.
Example for sum:
$$(3 \cdot 7^0+4 \cdot 7^1+2 \cdot 7^2)+(5 \cdot 7^0+3 \cdot 7^1)= \\ (3+5) \cdot 7^0+(4+3) \cdot7^1+2 \cdot 7^2= \\ (7+1) \cdot 7^0+7 \cdot 7^1+2 \cdot 7^2= \\ 1\cdot 7^0+(7+1) \cdot 7^1+2 \cdot 7^2= \\ 1 \cdot 7^0+1 \cdot 7^1+(2+1) \cdot 7^2=\\ 1 \cdot 7^0+1 \cdot 7^1+3 \cdot 7^2$$
An other method for the calculation:
(we start from the left side)

Example for multiplication:
$$(3 \cdot 7^0+4 \cdot 7^1+2 \cdot 7^2) \cdot (5 \cdot 7^0+3 \cdot 7^1)= \\ (3 \cdot5) \cdot 7^0+(4 \cdot 5+3 \cdot 3) \cdot7^1+(4 \cdot 3+2 \cdot 5) \cdot 7^2+(2 \cdot 3) \cdot 7^3= \\ 15 \cdot 7^0+29 \cdot 7^1+22 \cdot 7^2+6 \cdot 7^3= \\ (2 \cdot 7+1)\cdot 7^0+(4 \cdot7+1) \cdot 7^1+(3 \cdot 7+1) \cdot 7^2+6 \cdot 7^3= \\ 1 \cdot 7^0+(2+1) \cdot 7^1+(4+1) \cdot 7^2+(6+3) \cdot 7^3=\\ 1 \cdot 7^0+3 \cdot 7^1+5 \cdot 7^2+2 \cdot 7^3+1 \cdot 7^4$$
An other method for the calculation:
(we start from the left side)

I have not understand the other method for the multiplication.
Any help would be appreciated!
 [/INST] The other method for multiplication is a reflected version of "long multiplication".
You'll forgive me if I start on the left with the highest power of 7 (in your example) and align on the right, because what we're looking at here is working arithmetic in base 7.  But that just means you need to read the other way across the page.
The multiplication separates one of the multipliers into its component powers, does two simpler multiplications then adds the results:
$$(2 \cdot 7^2+4 \cdot 7^1+3 \cdot 7^0)\cdot \overbrace{(3 \cdot 7^1+5 \cdot 7^0)}^\text{separate this}= \hspace{2in} \\ 
(2 \cdot 7^2+4 \cdot 7^1+3 \cdot 7^0)\cdot (3 \cdot 7^1) \hspace{1in} \\
\hspace{1in} +(2 \cdot 7^2+4 \cdot 7^1+3 \cdot 7^0)\cdot (5 \cdot 7^0)  \\
\begin{array}{rrrr}
\hline
=&&(2\cdot 3 ) 7^3 &+(4\cdot 3) 7^2 &+(3\cdot 3) 7^1&\\
+&& &(2\cdot 5) 7^2 &+(4\cdot 5) 7^1&+(3\cdot 5)7^0\\ 
\hline
=&(1 ) 7^4&~_1+(0 ) 7^3 &~_1+(6) 7^2 &~_1+(2) 7^1&\\
+&& (1 ) 7^3&~_1+(6) 7^2 &~_3+(1) 7^1&~_2+(1)7^0\\ 
\hline
=&(1)7^4&+(2) 7^3 &+(5) 7^2 &+(3) 7^1&+(1)7^0\\
\end{array}$$
The multiplications of each component are resolved in my version right to left. The small subscripts are the multiple of 7 taken out of the lower-power result to their right and to be added on to the results of the multiplication for the next higher power on their left - the "carries". 

Note in particular that the numbers in brackets on the last few lines correspond exactly to (a reflected version of) the numbers in the layout that you had problems with.
</s>