#Java 10, (削除) 212 (削除ここまで)(削除) 211 (削除ここまで)(削除) 196 (削除ここまで)(削除) 195 (削除ここまで)(削除) 146 (削除ここまで) 143 bytes

Java 10, (削除) 212 (削除ここまで)(削除) 211 (削除ここまで)(削除) 196 (削除ここまで)(削除) 195 (削除ここまで)(削除) 146 (削除ここまで) 143 bytes

#Java 10, (削除) 212 (削除ここまで) (削除) 211 (削除ここまで) (削除) 196 (削除ここまで) (削除) 195 (削除ここまで) 146 bytes

s->n->{String r="",t;for(int m=0,c,j,i=0;i<s.length()-n;i++)for(c=0,j=-1;(j=s.indexOf(t=s.substring(i,i+n),j+1))>=0;)if(++c>m){m=c;r=t;}return r;}

-15 bytes by outputting the substrings with their count, which is allowed.
-1 byte thanks to @ceilingcat.
-49 bytes thanks to @OlivierGrégoire.

Try it online.

s->n->{ // Method with String and Integer parameters and String return-type
 String r="", // Result-String, starting empty
 t; // Temp-String
 for(int m=0, // Largest count, starting at 0
 c, // Temp count integer
 j, // Temp index integer
 i // Index-integer `i`
 =0;i<s.length()-n;i++)
 // Loop `i` in the range [0, length - n):
 for(c=0, // Reset the temp-count to 0
 j=-1; // And the temp-index integer to -1
 (j=s.indexOf(t=s.substring(i,i+n)
 // Set String `t` to a substring of the input-String in the range [i,i+n)
 j+1))// Set the temp-index `j` to the index of String `t` in the input-String,
 // only looking at the trailing portion after index `j+1`
 >=0;) // And continue looping as long as long as that substring is found
 if(++c>m){// If the temp-count + 1 is larger than the current largest count:
 m=c; // Set this current largest count to this temp-count + 1
 r=t;} // And the result-String to this temp-String
 return r;} // After the nested loops, return the result-String

#Java 10, (削除) 212 (削除ここまで) (削除) 211 (削除ここまで) (削除) 196 (削除ここまで) (削除) 195 (削除ここまで) (削除) 146 (削除ここまで) 143 bytes

s->n->{String r="",t;for(int m=0,c,j,i=s.length();i-->n;)for(c=0,j=-1;(j=s.indexOf(t=s.substring(i-n,i),j+1))>=0;)if(++c>m){m=c;r=t;}return r;}

-15 bytes by outputting the substrings with their count, which is allowed.
-1 byte thanks to @ceilingcat.
-52 bytes thanks to @OlivierGrégoire.

Try it online.

s->n->{ // Method with String and Integer parameters and String return-type
 String r="", // Result-String, starting empty
 t; // Temp-String
 for(int m=0, // Largest count, starting at 0
 c, // Temp count integer
 j, // Temp index integer
 i // Index-integer `i`
 =s.length();i-->n;)
 // Loop `i` in the range (input-length, n]:
 for(c=0, // Reset the temp-count to 0
 j=-1; // And the temp-index integer to -1
 (j=s.indexOf(t=s.substring(i-n,i)
 // Set String `t` to a substring of the input-String in the range [i-n,i)
 j+1))// Set the temp-index `j` to the index of String `t` in the input-String,
 // only looking at the trailing portion after index `j+1`
 >=0;) // And continue looping as long as long as that substring is found
 if(++c>m){// If the temp-count + 1 is larger than the current largest count:
 m=c; // Set this current largest count to this temp-count + 1
 r=t;} // And the result-String to this temp-String
 return r;} // After the nested loops, return the result-String

#Java 10, (削除) 212 (削除ここまで) (削除) 211 (削除ここまで) 196 bytes

import java.util.*;s->n->{var m=new HashMap<String,Long>();for(int l=s.length()-n+1;l-->0;)m.merge(s.substring(l,l+n),1L,Long::sum);return m.entrySet().stream().max(Map.Entry.comparingByValue());}

-15 bytes by outputting the substrings with their count, which is allowed.

Try it online.

import java.util.*; // Required import for HashMap and Map
s->n->{ // Method with String and Integer parameters and Optional
 // String-Integer Key-Value pair return-type
 var m=new HashMap<String,Long>();
 // A key-value map for String-Long pairs for the count per substring
 for(int l=s.length()-n+1;l-->0;)
 // Loop `l` in the range (input_length-n+1, 0]:
 m.merge(s.substring(l,l+n),
 // Get the substring in the index-range [l, l+n) as key
 1,Long::sum);
 // And add it to the map by either starting at 1, or adding 1,
 // depending on whether the key-value pair already exists or not
 return m.entrySet() // Then convert the Map to a Set of Key-Value pairs
 .stream() // Convert that Set to a Stream
 .max( // Get the maximum of that stream by:
 Map.Entry.comparingByValue())
 // Looking at the values (the counts)
 // And return this Optional key-value pair as result

#Java 10, (削除) 212 (削除ここまで) (削除) 211 (削除ここまで) (削除) 196 (削除ここまで)(削除) 195 (削除ここまで) 146 bytes

s->n->{String r="",t;for(int m=0,c,j,i=0;i<s.length()-n;i++)for(c=0,j=-1;(j=s.indexOf(t=s.substring(i,i+n),j+1))>=0;)if(++c>m){m=c;r=t;}return r;}

-15 bytes by outputting the substrings with their count, which is allowed .
-1 byte thanks to @ceilingcat.
-49 bytes thanks to @OlivierGrégoire.

Try it online.

s->n->{ // Method with String and Integer parameters and String return-type
 String r="", // Result-String, starting empty
 t; // Temp-String
 for(int m=0, // Largest count, starting at 0
 c, // Temp count integer
 j, // Temp index integer
 i // Index-integer `i`
  =0;i<s.length()-n;i++)
 // Loop `i` in the range [0, length - n):
 for(c=0, // Reset the temp-count to 0
 j=-1; // And the temp-index integer to -1
 (j=s.indexOf(t=s.substring(i,i+n)
 // Set String `t` to a substring of the input-String in the range [i,i+n)
 j+1))// Set the temp-index `j` to the index of String `t` in the input-String,
 // only looking at the trailing portion after index `j+1`
 >=0;) // And continue looping as long as long as that substring is found
 if(++c>m){// If the temp-count + 1 is larger than the current largest count:
 m=c; // Set this current largest count to this temp-count + 1
 r=t;}  // And the result-String to this temp-String
 return r;} // After the nested loops, return the result-String