Why ratio of photon energies at two locations are not given by ratio of metric values at those locations? What I know
I am told that in an inertial frame, the 4-momentum of a photon is:
$$p^{\mu}=\frac{E}{c}\left(1, \frac{\vec{p}}{|\vec{p}|}\right).\tag{1}$$
I can also show with the geodesic equation, that for a stationary metric, $p_0$ is constant.

Physical Setup
Lets have a photon emitted from the surface of a gravitating object. It is received by an observer at infinity. I am interested in how much the photon has redshifted.
To get the energy of the photon observed by the observer, I do, using the above equation:
$$E_{\text{Obs}}=p^0_{\text{Obs}}c=g^{00}_{\text{Obs}} p_0c\tag{2}$$
Where I have raised the index with the metric: $p^0=g^{00}p_0$.
Let's have on observer where the photon was emitted. It'll detect an energy for it:
$$E_{\text{Emit}}=p^0_{\text{Emit}}c=g^{00}_{\text{Emit}} p_0c\tag{3}$$.
To get ratio of frequencies, I divide the energies. $p_0$ and $c$ is the same for both observers. I obtain:
$$\frac{E_{\text{Obs}}}{E_{\text{Emit}}}=\frac{g^{00}_{\text{Obs}}}{g^{00}_{\text{Emit}}}=\frac{g_{00 \text{Emit}}}{g_{00 \text{Obs}}}\tag{4}$$

The problem
When I compare with my lecture notes, I find out that I am missing a square root & I should have obtained:
$$\frac{E_{\text{Obs}}}{E_{\text{Emit}}}=\sqrt{\frac{g_{00 \text{Emit}}}{g_{00 \text{Obs}}}}\tag{5}$$
What is wrong with my above calculation?
 A: In order to measure the frequency of a photon one needs an observer and its 4-velocity which must be properly normalized with the metric. A static observer will have $v^\mu = (v^0, 0, 0, 0)$ normalized with $g_{\mu\nu} v^\mu v^\nu = g_{00} (v^0)^2 = 1$ which gives $v^0 = 1/\sqrt{g_{00}}$.
If the spacetime is stationary it has a Killing-Vector $\xi^\mu = (1,0,0,0)$.
Let the photon has a 4-velocity $u^\mu$ normalized to zero $u_\mu u^\mu = 0$ along its geodesics.
Then it has a constant of motion $\omega = u_\mu \xi^\mu = u_0$ which is its energy.
The frequency (energy) measured by the observer $v^\mu$ is $E_{obs} = u_\mu v^\mu = v^0 = 1/\sqrt{g_{00}}$.
That is why the ratio of two frequencies measured at different positions is given by the ratio of the inverse(!) square roots of the metric component $g_{00}$:
$$ \frac{E_2}{E_1} = \frac{\sqrt{g_{00}(x_1)}}{\sqrt{g_{00}(x_2)}} $$
This is the basic formula giving the well-known redshift in the Schwarzschild metric.
Referring to your calculation: I guess you got the normalization of $v^\mu$ wrong.  Neither $v_0$ nor $v^0$ is constant and independent of the metric which is what you probably assumed. (I set $c=1$ for convenience but this should not be an issue to complement it.)
