I made some changes to your code. First, the "buff" variable needs the three level format (max number of characters allowed, another byte for the number of characteres entered, and the buffer itself) because that's what service 0AH requires. To use service 0AH I added "offset buff" (as Wolfgang said). Here it is:
.model small
.stack 100h
.data
buff db 26 ;MAX NUMBER OF CHARACTERS ALLOWED (25).
db ? ;NUMBER OF CHARACTERS ENTERED BY USER.
db 26 dup(0) ;CHARACTERS ENTERED BY USER.
.code
main:
mov ax, @data
mov ds, ax
;CAPTURE STRING FROM KEYBOARD.
mov ah, 0Ah ;SERVICE TO CAPTURE STRING FROM KEYBOARD.
mov dx, offset buff
int 21h
;CHANGE CHR(13) BY '$'.
mov si, offset buff + 1 ;NUMBER OF CHARACTERS ENTERED.
mov cl, [ si ] ;MOVE LENGTH TO CL.
mov ch, 0 ;CLEAR CH TO USE CX.
inc cx ;TO REACH CHR(13).
add si, cx ;NOW SI POINTS TO CHR(13).
mov al, '$'
mov [ si ], al ;REPLACE CHR(13) BY '$'.
;DISPLAY STRING.
mov ah, 9 ;SERVICE TO DISPLAY STRING.
mov dx, offset buff + 2 ;MUST END WITH '$'.
int 21h
mov ah, 4ch
int 21h
end main
When 0AH captures the string from keyboard, it ends with ENTER (character 13), that's why, if you want to capture 25 characters, you must specify 26.
To know how many characters the user entered (length), access the second byte (offset buff + 1). The ENTER is not included, so, if user types 8 characters and ENTER, this second byte will contain the number 8, not 9.
The entered characters start at offset buff + 2, and they end when character 13 appears. We use this to add the length to buff+2 + 1 to replace chr(13) by '$'. Now we can display the string.
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