How are variable values stored in C?

Your understanding is completely wrong.

When you write int x = 4;, the x represents an actual memory location on the stack, that then gets filled with the value 4. x is irrevocably linked with that piece of memory - when x goes out of scope the memory also disappears.

When you write int y = x; again y represents an actual piece of memory. It does not 'refer' to x, instead, the contents of x are copied into y.

Is it the same for all languages?

No, different languages can and do have completely different semantics. However the way C does it is usually called value semantics.

y never references x. The assignment operator, =, copies values. x is just a value of 4, of int type. int y = x is assigning the current value of x to y, copying it in the process.

To behave like you're describing, y would need to be a pointer to an int, int *, and it would be assigned the address of x, like so:

#include <stdio.h>

int main(int argc, char *argv[]) {
    int x = 4;
    int *y = &x;
    
    
    printf("before: x: %d, y: %d\n", x, *y);
    
    x = 123; // modify x
    
    printf("after:  x: %d, y: %d\n", x, *y);
}

X now references the memory location where the '4' is stored

No, 4 isn't stored anywhere, it's a parameter to a mov. x has its own memory location that holds an integer value, in this case 4.

y references x

No, y also has its own memory location that stores an integer, also in this case 4.

So why is it that when I change the value of x, eg x=6;, y doesn't get changed

They're both different memory locations, changing one has no impact on the other.