f(i,m) denotes number of substrings start at i and have m mod under 3. So, lets say we have a substring i.......k which has m mod under 3. So we can write,

(str[i] + str[i+1] + ...... + str[k])%3 = m%3

or

(str[i+1]+ ........ + str[k])%3 = (m — str[i])%3

So for solving f(i,m) if we remove first position letter the remaining substring should have mod = (m — x)%3. So, how come it becomes (m+x)%3. I even checked with (m-x)%3 code and it works as well:

long long f(int i, int m){
    if(i == (int)s.size()){
        return 0; 
    }
    if(memoize[i][m] != -1) {
        return memoize[i][m]; 
	}
    int x=s[i]-'0'; 
    int md = (m-x)%3;
    if(md<0) md += 3;
    int ans = f(i + 1, md) + (md == 0 and x % 2 == 0); 

    return memoize[i][m] = ans;    
}