Your result (array3) must also be a Uint8Array array.

So, in your example the following should achieve what you're looking for:

const array3 = new Uint8Array([ ...array1, ...array2]);

You can do:

var array2 = Array.from(uint8Array);
var array3 = array1.concat(array2);

Your code is creating a new array with your four strings followed by the Uint8Array as a single element.

Array.prototype.concat is an odd beast. If you give it arguments that aren't "concat spreadable" (for instance, things that aren't arrays), it includes those values as simple elements in the array it creates.

Typed arrays are not concat spreadable:

const uint8 = Uint8Array.from([3, 4]);
console.log(uint8[Symbol.isConcatSpreadable]); // undefined (and the default is false)
const softArray = [1, 2];
const result = softArray.concat(uint8);
console.log(result.length); // 3
console.log(result[0]); // 1
console.log(result[1]); // 2
console.log(result[2]); // (A Uint8Array containing 3, 4)

You could use spread instead of concat:

const array1 = ['B',  'G',  'R',  'A'];
const array2 = Uint8Array.from([1, 2, 3]);
const result = [...array1, ...array2];
console.log(result); // ["B", "g", "R", "a", 1, 2, 3]