Can anyone explain this Javascript anomaly with the Map function?

The member accessor . has higher precedence than +

const obj = { foo: 41 };

console.log( 1 + obj.foo );   // 42
console.log( 1 + (obj.foo) ); // 42
console.log( (1 + obj).foo ); // undefined

Thus in 131+"".split(',').map(Number) the order of execution is (slightly abbreviated to focus on the important parts):

    131 + "".split(',') .map(Number)
//      ^ ^^^^^^^^^^^^^  ^^^^^^^^^^^ 
//      3       1             2

1. "".split(',') - split the empty string using a single comma. This results in [ "" ] (see The confusion about the split() function of JavaScript with an empty string):

   console.log( "".split(',') ); // [ "" ]

2. [ "" ].map(Number) - the Number function is applied to each member of the array. Since there the array after the split has a single empty string, it results in [ 0 ] because Number("") is 0 (see Number called using empty string, null, or undefined):

   console.log( Number("") );             // 0
   console.log( [ "" ].map(Number) ); // [ 0 ]

3. 131 + [ 0 ] - adding a number and an array. This will first type coerce the array into a primitive then based on what kind of primitive it is (string or number) it will perform either numeric addition with 131 or string concatenation (coercing the other operand to a string first). Arrays convert to a string primitive by joining all members together (see Why is [1,2] + [3,4] = "1,23,4" in JavaScript?). As the array only has a single member (a zero) the result is that single member as a string.

   const result = 131 + [ 0 ];
   console.log(result, typeof result);                // 1310 string
   
   const arrayCoercion = "" + [ 0 ];
   console.log(arrayCoercion , typeof arrayCoercion); // 0 string

Therefore the code concatenates 131 and 0 together.