What is exactly the base pointer and stack pointer? To what do they point?

ESP is the current stack pointer, which will change any time a word or address is pushed or popped onto/off off the stack. EBP is a more convenient way for the compiler to keep track of a function's parameters and local variables than using the ESP directly.

Generally (and this may vary from compiler to compiler), all of the arguments to a function being called are pushed onto the stack by the calling function (usually in the reverse order that they're declared in the function prototype, but this varies). Then the function is called, which pushes the return address (EIP) onto the stack.

Upon entry to the function, the old EBP value is pushed onto the stack and EBP is set to the value of ESP. Then the ESP is decremented (because the stack grows downward in memory) to allocate space for the function's local variables and temporaries. From that point on, during the execution of the function, the arguments to the function are located on the stack at positive offsets from EBP (because they were pushed prior to the function call), and the local variables are located at negative offsets from EBP (because they were allocated on the stack after the function entry). That's why the EBP is called the Frame Pointer, because it points to the center of the function call frame.

Upon exit, all the function has to do is set ESP to the value of EBP (which deallocates the local variables from the stack, and exposes the entry EBP on the top of the stack), then pop the old EBP value from the stack, and then the function returns (popping the return address into EIP).

Upon returning back to the calling function, it can then increment ESP in order to remove the function arguments it pushed onto the stack just prior to calling the other function. At this point, the stack is back in the same state it was in prior to invoking the called function.

esp is as you say it is, the top of the stack.

ebp is usually set to esp at the start of the function. Function parameters and local variables are accessed by adding and subtracting, respectively, a constant offset from ebp. All x86 calling conventions define ebp as being preserved across function calls. ebp itself actually points to the previous frame's base pointer, which enables stack walking in a debugger and viewing other frames local variables to work.

Most function prologs look something like:

push ebp      ; Preserve current frame pointer
mov ebp, esp  ; Create new frame pointer pointing to current stack top
sub esp, 20   ; allocate 20 bytes worth of locals on stack.

Then later in the function you may have code like (presuming both local variables are 4 bytes)

mov [ebp-4], eax    ; Store eax in first local
mov ebx, [ebp - 8]  ; Load ebx from second local

FPO or frame pointer omission optimization which you can enable will actually eliminate this and use ebp as another register and access locals directly off of esp, but this makes debugging a bit more difficult since the debugger can no longer directly access the stack frames of earlier function calls.

EDIT:

For your updated question, the missing two entries in the stack are:

var_C = dword ptr -0Ch
var_8 = dword ptr -8
var_4 = dword ptr -4
*savedFramePointer = dword ptr 0*
*return address = dword ptr 4*
hInstance = dword ptr  8h
PrevInstance = dword ptr  0C
hlpCmdLine = dword ptr  10h
nShowCmd = dword ptr  14h

This is because the flow of the function call is:

• Push parameters (hInstance, etc.)
• Call function, which pushes return address
• Push ebp
• Allocate space for locals

You have it right. The stack pointer points to the top item on the stack and the base pointer points to the "previous" top of the stack before the function was called.

When you call a function, any local variable will be stored on the stack and the stack pointer will be incremented. When you return from the function, all the local variables on the stack go out of scope. You do this by setting the stack pointer back to the base pointer (which was the "previous" top before the function call).

Doing memory allocation this way is very, very fast and efficient.