Which formula to use to find the pH of a solution?
From reading your question it is obvious that you are confused. When I studied chemistry in college in the 1970's I don't remember the Henderson-Hasselbach equation being a famous equation. It seems like it is now being used as some magic can opener.

You need to go back to the chemistry and figure out the chemistry first, then do the math. 

Ok, let's start by noting that the pKa of acetic acid is 4.76 from Wikipedia. There are two significant figures in the mantissa so a mathematical solution good to 1% will be adequate. Also we are dealing with solutions on the order of $10^{-3}$ molar. This allows us to make two very important assumptions which significantly simply the mathematics.


First we assume that we can ignore the autoionization of water. 
Second we assume that since acetic acid is a weak acid the solution ends up being essentially 0.004 molar sodium acetate and 0.001 molar acetic acid.  


Because we have excess acetic acid we know that the solution will be acidic. If we wanted a more exact solution we'd ignore the second assumption and use the $K_\mathrm{a}$ expression 

$K_\mathrm{a} = \dfrac{\ce{[H^+][Ac^-]}}{\ce{[HAc]}}$

to solve the problem more exactly. Letting $\ce{[H^+]} = x$ then $\ce{[Ac^-]} = 0.004 + x$ and $\ce{[HAc^-]} = 0.001 - x$. This requires solving a quadratic equation which is doable, but mathematically messier.   

Now back to using our two assumptions we can go to the Henderson–Hasselbalch equation and solve the problem with a simple formula. 

$\mathrm{pH} = \mathrm{pKa} + \log{\dfrac{\ce{[Ac^-]}}{\ce{[HAc]}}} = 4.76 +  \log{\dfrac{0.004}{0.001}} = 5.36$

This means that the $\ce{[H^+]} = 4.4 \times 10^{-6}$. 


So our assumption that the final solution ends up being essentially 0.004 molar sodium acetate and 0.001 molar acetic acid is reasonable.  
The assumption that the autocatalysis of water can be ignored is a bit dicier.