Understanding how the copying of the compressed Linux kernel works in head_64.S

I am working on understanding how the Linux kernel v3.19 prepares for decompression on a machine with x86_64 architecture in arch/x86/boot/compressed/head_64.S. The part that is confusing me is the copying of the compressed kernel image to another location in memory before decompression takes place. I have copied the code below with my understanding of each line; the lines with // ? comments are the ones I am having trouble with.

/*
 * Copy the compressed kernel to the end of our buffer
 * where decompression in place becomes safe.
 */
    pushq   %rsi  // %rsi contains the pointer to `boot_params`
    leaq    (_bss-8)(%rip), %rsi  // ?
    leaq    (_bss-8)(%rbx), %rdi  // ?
    movq    $_bss /* - $startup_32 */, %rcx  // ?
    shrq    $3, %rcx  // ?
    std           // set the direction so %rsi and %edi are decremented
    rep movsq     // copy data from %rsi to %rdi 8 bytes at a time
    cld           // clear the direction flag
    popq    %rsi  // restore pointer to `boot_params`

I know that at this point %rbx contains the location where the kernel will be relocated for decompression and that %rbp (while not mentioned in this code snippet) contains the location the kernel was loaded at runtime. Further, I am aware that _bss refers to the start of the .bss section as given by the linker script in arch/x86/boot/compressed/vmlinux.lds.S, and that the compressed kernel image that we want to copy is at .rodata..compressed, which sits between .head.text and .text (where we are currently, I believe).

What I am not following and need help with is:

1. What section does _bss-8 refer to?

2. This may become clear once I understand what _bss-8 is, but how does (_bss-8)(%rip) and (_bss-8)(%rbx) give us where the compressed kernel currently is and where it is going?

3. Why are we putting $_bss >> 3 into %rcx as the counter for copying data? Why not something related to rodata..compressed?