Main Question
The main correction is due to the gravitational effects
of the Earth clock being in the potential well of the Earth, followed by the effects of the Moon clock being in the potential well of the Moon and of the Earth, and the effects of the relative speeds of the clocks are only felt in the third significant figure. I obtain an overall correction of 56.0 $\mu$s/day and cannot reproduce the statement value of 58.7 $\mu$s/year unless ignoring the fact that the Moon clock is in the potential well of the Moon.
Details
An answer can be provided in terms of the Schwarzschild metric interval, which says
$$c^2 d\tau^2 = \left(1 - \frac{r_s}{r}\right)c^2dt^2 - \left(1 - \frac{r_s}{r}\right)^{-1} dr^2 - r^2d\theta^2 - r^2\sin^2\theta d\phi^2$$
in the usual Schwarzschild-Droste coordinates, where $d\tau$ is a proper time interval and $r_s$ is the Schwarzschild radius.
If we set the Moon's orbital plane to be fixed at $\theta=\pi/2$ ($d\theta=0$) and assume $r_s/d_m \ll 1$, then an object in a Moon-like orbit around the Earth, at the distance of the Moon, $d_m$ will satisfy
$$ \frac{d\tau_{m,o}}{dt} = \left[1 - \frac{r_{s,e}}{d_m} - \frac{v_m^2}{c^2} \right]^{1/2}\ ,$$
where $v_m$ is the orbital speed of the Moon and $r_{s,e}$ is the Schwarzschild radius of the Earth.
However, a clock on the Moon's surface will run slower than this because of the Moon's gravitational potential that can also be represented with the Schwarzschild metric (assuming the gravitational influence of the Earth can be neglected at the surface of the Moon - i.e. $M_m/r_m^2 \gg M_e/d_m^2$).
$$\frac{d\tau_{m,s}}{d\tau_{m,o}} = \left[1 - \frac{r_{s,m}}{r_m}\right]^{1/2}\ ,$$
where $r_{s,m}$ is the Schwarzschild radius of the Moon and we assume the Moon's surface moves at the same velocity as something in orbit around the Earth at the distance of the Moon.
Finally, a clock on the Earth's surface has its own proper time given by
$$\frac{d\tau_{e,s}}{dt} = \left[1 - \frac{r_{s,e}}{r_e} - \frac{v_e^2}{c^2} \right]^{1/2}\ , $$
where $v_e$ is the speed of the clock on the Earth's surface with respect to the centre of the Earth.
Putting this together, the ratio of proper time intervals for a clock on the surface of the Moon and a clock on the surface of the Earth is
$$\frac{d\tau_{m,s}}{d\tau_{e,s}} = \frac{d\tau_{m,s}}{d\tau_{m,o}} \frac{d\tau_{m,o}}{dt} \frac{dt}{d\tau_{e,s}} = \left[1 - \frac{r_{s,m}}{r_m}\right]^{1/2} \left[1 - \frac{r_{s,e}}{d_m} - \frac{v_m^2}{c^2} \right]^{1/2} \left[1 - \frac{r_{s,e}}{r_e} - \frac{v_e^2}{c^2} \right]^{-1/2}\ .$$
All the terms $r_{s,m}/r_m$, $r_{s,e}/d_m$, $r_{s,e}/r_e$, $v_m^2/c^2$ and $v_e^2/c^2$ are $\ll 1$, so binomial expansions can be employed, keeping only terms at this order.
$$\frac{d\tau_{m,s}}{d\tau_{e,s}} = 1 - \frac{1}{2}\left[ \frac{r_{s,m}}{r_m} + \frac{r_{s,e}}{d_m} + \frac{v_m^2}{c^2} - \frac{r_{s,e}}{r_e} - \frac{v_e^2}{c^2}\right]\ . $$
We can now put in numbers and explicitly write down the numerical values of each of the correction terms to see which are the most important. I have used $r_{s,e} = 8.869\times 10^{-3}$ m, $r_{s,m} = 1.09\times 10^{-4}$ m, then assuming clocks on the equators of the Moon and Earth, so $r_e = 6.378 \times 10^6$ m, $r_m = 1.74\times 10^6$ m. For $v_e$, I assume $0.465\times 10^3$ m/s, for a mean orbital speed I assume $v_m = 1.02\times 10^3$ m/s and for an average Earth-Moon distance, $d_m = 3.85 \times 10^8$ m.
$$\frac{d\tau_{m,s}}{d\tau_{e,s}} = 1 - \frac{1}{2}\left[ 6.26\times 10^{-11} + 2.30\times 10^{-11} + 1.16\times 10^{-11} - 1.391\times 10^{-9} - 2.40\times 10^{-12}\right] $$
Thus the most important term is due to a clock on the Earth in the deep gravitational potential well of the Earth; the next most important is the clock on the Moon being in the potential well of the Moon. Then, that the clock on the Moon is still in the potential well of the Earth and the speed terms make contributions to the third significant figure.
In summary
$$\frac{d\tau_{m,s}}{d\tau_{e,s}} = 1 - 6.48\times 10^{-10}\ .$$
Multiplying the correction term by the number of seconds in 24 hours, we have that a clock on the Moon runs faster by 56.0 $\mu$s/day.
Bonus Question
There will then be periodic terms on the period of the Moon's orbit around the Earth. These will: (i) cause $d_m$ to change by about $\pm 5$%, which, from the contribution of the second term, leads to variations of up to $\pm 0.9$ $\mu$s/day. (ii) There is also a contribution from the Earth and Moon sitting at different points in the Sun's gravitational potential by up to (approximately) the distance between the Earth and the Moon. This leads to a change in the relative time dilation of up to
$$ \Delta \left(\frac{d\tau_{m,s}}{d\tau_{e,s}}\right) \simeq \frac{1}{2} \left[ \frac{r_{s,\odot}}{d_e} - \frac{r_{s,\odot}}{d_e + d_m}\right] \simeq \frac{d_m r_{s,\odot}}{2d_e^2}\ ,$$
where $d_e$ is the Earth-Sun distance and $r_{s,\odot}$ is the Schwarzschild radius of the Sun. Using $d_e = 1.50\times 10^{11}$ m and $r_{s,\odot}=2.96\times 10^3$ m yields a periodic correction of up to $\pm 2.2$ $\mu$s/day.
Coda
I cannot reproduce the statement value of $58.7$ $\mu$s/day. I note however, that this is what I get if I ignore the correction for the clock being on the surface of the Moon... i.e. I think that the figure of $58.7$ $\mu$s/day is the correction for a clock in orbit around the Earth at the distance of the Moon, not a clock on the Moon's surface.
