Because of the expansion of space itself, did the light from a galaxy one million light-years away actually take more than one million years to reach us?
Yes, that's correct. The proper distance to an object at the time of observation ($d_p(t_0)$) is related to the proper distance when the light was emitted ($d_p(t_e)$) via
$$d_p(t_0) = d_p(t_e) \times (1+z)~~Eqn.(1)$$
For light $ds=0$ and by taking $d\Omega = 0$, the FRWL metric (for flat universe)
$$ds^2 = -c^2dt^2 + a(t)^2[dr^2 + r^2d\Omega^2]$$
takes the form of
$$r = c\int_{t_e}^{t_o}\frac{dt}{a(t)}$$
But you can write this equation in terms of z to make calculations easier.
So we know that $$1 + z = a(t)^{-1}$$
hence
$$\frac{dz}{dt} = \frac{dz} {da} \frac{da} {dt}$$
$$\frac{dz}{dt} = -\frac{1}{a^2} \dot{a}$$
$$dz = -\frac{H}{a}dt = -H(1+z)dt$$
$$dt = -\frac{dz}{H(1+z)}$$
Since at $t_e$ corrsponds to $z$ and $t_0 = 0$ we have
$$r = c\int_{t_e}^{t_o}\frac{dt}{a(t)} = -c\int_{z}^{0}\frac{dz}{H}$$
Also we know that $$H = H_0\sqrt{\Omega_r(1+z)^4 + \Omega_m(1+z)^3 + \Omega_{\Lambda}}$$
Finally
$$ r = -\frac{c}{H_0}\int_{z}^{0}\frac{dz}{\sqrt{\Omega_r(1+z)^4 + \Omega_m(1+z)^3 + \Omega_{\Lambda}}}$$
or $$ r = \frac{c}{H_0}\int_{0}^{z}\frac{dz}{\sqrt{\Omega_r(1+z)^4 + \Omega_m(1+z)^3 + \Omega_{\Lambda}}}~~Eqn.(2)$$
Also, if we are looking at a galaxy one billion light-years away, how far away is it today?
If you know the $z$ value you can easily calculate the desired values by using Equation (1) and Equation (2)
I understand (I think) that if an astronomer sees light from 13.8 billion light-years away, it is now 46.6 light-years away, a ratio of about 3.375...
You can easily calculate this factor by using Equation 2. We are looking for the 13.8 billion light-years away so in other terms the beginning of the universe. This implies that in Equation (2) the upper limit of the integral should be infinity.
$$ r = \frac{c}{H_0}\int_{0}^{\infty}\frac{dz}{\sqrt{\Omega_r(1+z)^4 + \Omega_m(1+z)^3 + \Omega_{\Lambda}}}$$
When you set $\Omega_{r} = 0$, $\Omega_{m} = 0.31$ and $\Omega_{\Lambda} = 0.69$ you ll find that ( you can use this site to do the integral)
$$r = \frac{c}{H_0}\int_{0}^{\infty}\frac{dz}{\sqrt{0.31(1+z)^3 + 0.69}} = 3.26133$$
$$r = 3.2 \times c/H_0 = 3.2 \times 4408.712 Mpc = 14107.88 Mpc = 46.013 Gly$$
Which is the desired result?
But, the ratio for something closer is going to be less, because of Hubble's law/parameter... Right?
It depends on what parameters you are using for $\Omega$'s. But I think you can easily plot a graph by taking the
$$r = \frac{c}{H_0}\int_{0}^{z}\frac{dz}{\sqrt{0.31(1+z)^3 + 0.69}}$$ and plotting for different z values.
