The answer to this question is addressed, for the case of a simple Schwarzschild black hole, by Taylor & Wheeler in their book called "Exploring black holes" (Addison, Wesley, Longman, 2000, pages B20-B24). There is a huge difference to what would be perceived by a "shell" observer that arranges to be stationary just outside the event horizon compared with an observer that is free-falling into the black hole.
Your question supposes that an observer is falling into the black hole. That observer will be able to see the outside universe from inside the event horizon. The light will be blue-shifted and it will be distorted/aberrated in such a way that as the observer approaches the singularity, the light from the rest of the universe is pushed outward (i.e. the viewing angle with respect to the fall-line becomes larger) into a halo and finally an intense ring of blue-shifted radiation that goes all around the sky, with blackness both in front of the observer and behind. Nothing special happens to the observer as they cross the event horizon.
A "shell" observer, using almost-infinite rocket power to hover just above the event horizon would see the whole universe compressed to a small, intense, blue-shifted dot overhead.
Of course there cannot be "shell" observers inside the event horizon since everything is compelled to move towards the singularity.
NB: This all just assumes classical GR theory. For further information and animations you could look at Andrew Hamilton's set of resources.
Write $a=1+\delta a$. Then (working to second order in $H_0^2 (t-t_0)^2$)
\begin{equation}
\delta a = H_0 (t-t_0) - \frac{q_0 H_0^2}{2} (t-t_0)^2
\end{equation}
So
\begin{eqnarray}
a^{-1} = \left(1+\delta a\right)^{-1} &=& 1 - \color{blue}{\delta a} + \color{red}{\delta a^2} + \cdots \\
&=& 1 - \color{blue}{\left[H_0 (t-t_0) - \frac{q_0 H_0^2}{2}(t-t_0)^2\right]} + \color{red}{\left[H_0^2 (t-t_0)^2 + \cdots \right]} + \cdots \\
&=& 1 - \color{blue}{H_0 (t-t_0)} + \left(\color{red}{1} + \color{blue}{\frac{q_0}{2}}\right)H_0^2(t-t_0)^2 + \cdots
\end{eqnarray}
where $+\cdots$ refers to terms that are order $H_0^3 (t-t_0)^3$ or higher.
In other words, it appears you left out the $\color{red}{\delta a^2}$ term when you did the Taylor expansion.
Best Answer
The orange line is a set of events all at the same cosmic time. The length of this line is indeed the proper distance between the brown and yellow lines, if we define "proper distance" to mean the distance on the spacelike surface of given cosmic time, and this is what is usually meant by the term "proper distance" in cosmology. (Another name for it is "ruler distance".)
The red line is a representation on the diagram of a null geodesic of the spacetime. The total integrated invariant interval along this line is therefore zero. The ends of the red line are at different comoving coordinate locations. The difference of those coordinate values provides information about the spatial separation of the emission and reception events, but to convert that information into a distance you have to do something a little artificial. You have to chose how to talk about distances between events at different times in a dynamic spacetime. There is no single prescription for that. You can say that a long time ago, when the light was emitted by the quasar, the proper distance between the Milky Way and the quasar was a lot smaller than it is in the present, and that proper distance can be obtained by using the metric at the emission time (you integrate along the spatial hypersurface, i.e. one of the lines of latitude (purple) on the diagram). Another way to give information about the distance here is to appeal to concepts such as "luminosity distance".
Sometimes people refer to emission events in the distant past and talk about distances to their comoving coordinate location now, in the present. That is a little misleading, it seems to me, but it is often going on when people talk about the size of "the observable universe".