Order of execution in this Assembly Code? (NASM)

I tried searching for an answer to this and the examples with any similarity at all were either too simple or too complex. I'm using Paul Carter's book for an assembly class, so some of the macros are from him. Namely, the print_string, print_int, and print_nl functions.

Given this snippet of assembly:

segment .data

    output db "Welcome!",0
    string1 db "greater than",0
    string2 db "less than",0
    string3 db "equal to",0
    var_a dd 0Ah

segment .bss

segment .text global _asm_main

_asm_main:

    enter   0,0               ; setup routine
    pusha
;***************CODE STARTS HERE***************************

    mov eax, output
    call print_string
    call print_nl

    cmp dword[var_a], 0Ah
    jle label1
    mov eax, string1
    call print_string
    call print_nl
    jmp label3

label1:
    cmp dword[var_a], 9h
    jg label2
    mov eax, string2
    call print_string
    call print_nl
    jmp label3
label2:
    mov eax, string3
    call print_string
    call print_nl
label3:
    cmp dword [var_a], 0
    jle label4
    sar [var_a], 1
    mov eax, dword [var_a]
    call print_int
    call print_nl
    jmp label3
label4:

;***************CODE ENDS HERE*****************************
    popa
    mov     eax, 0            ; return back to C
    leave
    ret

The output is:

Welcome!

equal to

5

2

1

0

My question:

I see how it eventually gets to label2 via the comparisons, prints the "equal to" message and a new line. After that though, I see no additional comparisons in label2. At that point, how does label3 even get invoked? How does the program "leave" label2? I understand everything in the assembly except in between the end of label2(which prints the "equal to") and the start of label3(Where it loops a bit shift and prints the numbers). Am I missing something?

After label2 is executed, does control of the program go back to where label2 was first called in label1? Or does label3 execute because it's next sequentially?