;Converts an int into a string
macro itoa int,destination,counter
{
mov eax,[int]
mov edi,destination

add edi,counter-1 ;vamos al final??
mov [edi+1],byte 0 ;0 ended strings!

mov ecx,counter ;number of loops
mov ebx,10
.loop:
xor edx,edx ;se necesita para la division
div ebx ;eax=number divided by ten. edx=modulo
add edx,'0' ;"0" is added to convert number to ascii char (0 + "0" = "0", 1+"0"="1", 5+"0"=5 etc.)
mov [edi],edx ;move the char to the string
dec edi ;continue with another char
loop .loop
;alternative way;
;cmp edi,destination
;jne .loop
 }             

macro itoa int,destination,counter
{
mov eax,[int]
mov edi,destination

add edi,[counter] 
dec edi
mov [edi+1],byte 0 ;0 ended strings!

mov ecx,[counter] ;number of loops
mov ebx,10
.loop:
xor edx,edx ;se necesita para la division
div ebx ;eax=number divided by ten. edx=modulo
add edx,'0' ;"0" is added to convert number to ascii char (0 + "0" = "0", 1+"0"="1", 5+"0"=5 etc.)
mov [edi],dl ;move the char to the string
dec edi ;continue with another char
loop .loop
;alternative way;
;cmp edi,destination
;jne .loop
 }    

mov [edi],edx ;move the char to the string     

mov [edi],dl ;move the char to the string 
    

itoa in_value,stringdestination,4    

;Converts an in into a string
macro itoa int,destination,counter
{
mov eax,[int]
mov edi,destination

add edi,counter;-1
dec edi
mov [edi+1],byte 0 ;0 ended strings!

mov ecx,counter ;number of loops
mov ebx,10
.loop:
xor edx,edx ;se necesita para la division
div ebx ;eax=number divided by ten. edx=modulo
add edx,'0' ;"0" is added to convert number to ascii char (0 + "0" = "0", 1+"0"="1", 5+"0"=5 etc.)
mov [edi],dl ;move the char to the string
dec edi ;continue with another char
loop .loop
;alternative way;
;cmp edi,destination
;jne .loop
 }      

but i found that in your code you used [counter]... and the correct use is counter without brackets... be cause the macro is called like this

lol, reinventing the wheel will never be easy

  mov esi,string
  mov edi,destinationinteger

  .loop2:
  sub esi,'0'

  mov eax,[edi]
  mov ebx,10
  mul ebx

  mov [edi],eax
  add [edi],esi
  inc esi
  loop .loop2    

To convert an ASCII string representation of a integer into the corresponding
integer.
Note that in this algorithm, the operation of getting the character at address X is
written as *X.
* Let S be a pointer to start of the string.
* Let D be the number.
1. Set D = 0.
2. Loop:
(a) If *S == '\n', then continue with the next step.
(b) Otherwise,
i. S = (S + 1)
ii. D = (D * 10)
iii. D = (D + (*S - '0'))
In this step, we can take advantage of the fact that ASCII puts the
numbers with represent the digits 0 through 9 are arranged consecutively,
starting at 0. Therefore, for any ASCII character x, the number
represented by x is simply x - '0'.

macro atoi{
mov esi , string
mov ecx , stringsize
mov edi , DestInt
mov edx , 0

local .loop
.loop:
    movzx eax , byte [esi]
    shl edx , 1                      ; <== edx = edx * 2
    sub eax , '0'
    lea edx , [edx * 4 + edx]  ; <== edx = edx * 5
    add edx , eax
    dec ecx
jnz .loop
}