Simple recursive function not giving expected output

Your code does the following (omitting the if):

countdown(n):
     print(n)
       countdown(n-1)
     print(n-1)

For n=2:

countdown(2):
     print(2)
       countdown(1)
     print(1)

-> next recursion step
     print(2)
       print(1)
         countdown(0)
       print(0)
     print(1)

-> next recursion step
     print(2)
       print(1)
         print(0)
         print(-1)
       print(0)
     print(1)

Your code has no problem. Just remove 2nd printf code.

void countdown(int n)
{
    printf("n = %d\t", n);
    n--;
    if (n >= 0)
        countdown(n);
}

int main()
{
    countdown(2);
    return 0;
}

The result is:

n = 2 n = 1 n = 0

This is the callstack when I captured at 2nd printf.

StudyCpp.exe!countdown(int n) line 16   C++  // It is 2nd printf of countdown(0). Now, n is -1. 
StudyCpp.exe!countdown(int n) line 14   C++  // It called countdown(0)
StudyCpp.exe!countdown(int n) line 14   C++  // It called countdown(1)
StudyCpp.exe!main() line 21 C++   // It called countdown(2)

If you progress debug one more, You can see the callstack as below:

StudyCpp.exe!countdown(int n) line 16   C++  // It is 2nd printf of countdown(1) after executed countdown(0).
StudyCpp.exe!countdown(int n) line 14   C++  // It called countdown(1)
StudyCpp.exe!main() line 21 C++   // It called countdown(2)

And, If you progress debug one more, You can see the callstack as below:

StudyCpp.exe!countdown(int n) line 16   C++  // It is 2nd printf of countdown(2) after executed countdown(1).
StudyCpp.exe!main() line 21 C++   // It called countdown(2)

And, Finally, Program will be exited.

Here you are calling countdown() recursively and is called thrice.
Actually each recursive call will push countdown() into the stack.

SF1(Bottom of stack) ---> SF2 ---> SF3 (Top of stack). 

Now the frame in the top will be executed.
During the return from the function, the particular stack frame will be popped out.
Now Stack frame pointer points to SF2 and then SF1.

Considering your program, below is the flow.
Push Operation:
SF1 will be pushed onto the stack first

n = 2 printed.  
Again n updated to 1

SF2 pushed:

n = 1 got printed.  
Again n updated to 0

SF3 pushed:

n = 0 got printed.  
Again n updated to -1.  
But n <= 0, So if check fails.  

Pop operation:
Now SF3 got popped out from stack first.

n = -1 printed  

Then SF2 popped

prints n = 0.

Finally SF1

n = 1 printed