You can create a new TypedArray with a new ArrayBuffer, but you can't change the size of an existing buffer

function concatTypedArrays(a, b) { // a, b TypedArray of same type
    var c = new (a.constructor)(a.length + b.length);
    c.set(a, 0);
    c.set(b, a.length);
    return c;
}

Now can do

var a = new Uint8Array(2),
    b = new Uint8Array(3);
a[0] = 1; a[1] = 2;
b[0] = 3; b[1] = 4;
concatTypedArrays(a, b); // [1, 2, 3, 4, 0] Uint8Array length 5

If you want to use different types, go via Uint8Array as the smallest unit is a byte, i.e.

function concatBuffers(a, b) {
    return concatTypedArrays(
        new Uint8Array(a.buffer || a), 
        new Uint8Array(b.buffer || b)
    ).buffer;
}

This means .length will work as expected, you could now convert this to your typed array of choice (make sure it's a type that would accept the .byteLength of the buffer though)

From here, you could now implement any method you like for concatenating your data, e.g.

function concatBytes(ui8a, byte) {
    var b = new Uint8Array(1);
    b[0] = byte;
    return concatTypedArrays(ui8a, b);
}

var u8 = new Uint8Array(0);
u8 = concatBytes(u8, 0x80); // [128]

You have to split your two byte number into two individual bytes. This can be done by bitwise shifting.

var twoBytes = 0xaaff;
var byte1 = ((twoBytes >> 8) & 0xff);
var byte2 = twoBytes & 0xff;
console.log(byte1);
console.log(byte2);

var buffer = new ArrayBuffer(2);
var Uint8View = new Uint8Array(buffer);
Uint8View[0] = byte1;
Uint8View[1] = byte2;

var Uint16View = new Uint16Array(buffer);
console.log(Uint16View[0]);