> What is a good mental representation of $dt$ vs $d\tau$?

I'm not quite sure I understand the question. Here's how I "mentally" interpret $t$.

$t$ is a coordinate. I wouldn't actually picture it as "time flowing", as I wouldn't picture the coordinate $r$ as "distance from the center". I would picture it simply as a coordinate, used to describe your world. So, in a sense, we may say it's common to all the observers. For example, instead of $(t,r,\theta,\phi)$ you may decide to use the Kruskal–Szekeres coordinates $(T,X, \theta, \phi)$, where here the time coordinate is $T=T(t,r)$. So, should we use $t$ or $T$, what is our "real" time, which is the best one? None of them. Depending on what physical situation you are analyzing, a set of coordinates may be more useful than another one, describing the physical context in a clearer way. Coordinates are a tool you are using to describe your world.

Let's focus on the Schwartzchild metric, as it's the most simple one:
\begin{equation}
ds^2=\Big(1-\frac{2m}{r}\Big)dt^2-\Big(1-\frac{2m}{r}\Big)^{-1}dr^2-r^2d\Omega^2
\end{equation}
which is asymptotically flat (i.e. we recover the Minkowski metric far away from the origin). In this case, $dt$ has a physical meaning asymptotically, as it represents the proper time measured by an observer at rest at $r\to\infty$. So we may say that $t$ represents a "real time" only for an observer at rest at infinity. 

Let's make an example to clarify the difference between $dt$ and $d\tau$ and how they are related. Let's consider a photon in a circular orbit ($r=3m$) in the plane $\theta=\pi/2$. This is a light-like event, so we have $ds^2=0$, thus computing from the Schwartzschild metric, we get
\begin{equation}
0=\frac{1}{3}dt^2-9m^2d\phi^2\implies dt^2=27m^2d\phi^2
\end{equation}
Now, say we want to know how much time passes for an observer who is at rest at $r=3m$. You can picture it this way: an observer at $r=3m$ "shoots" a photon with a laser and he starts a chronometer, and he stops it as he sees "his" photon again, once the photon has successfully completed the orbit. What we need is the relation
\begin{equation}
d\tau_{r=3m}=\frac{1}{3}dt^2
\end{equation}
obtained by setting $dr=d\phi=d\theta$ (as the observer is at rest) and $r=3m$. Substituting $dt^2=27m^2d\phi^2$ in the equation and integrating from $0$ to $2\pi$ we obtain $\Delta\tau_{r=3m}=6\pi m$. What would a distant observer at rest say about an event "taking" $dt^2=27m^2d\phi^2$ coordinate time? Again, from the Schwartzschild metric
\begin{equation}
d\tau_{r=\infty}=dt^2\implies \Delta\tau_{r=\infty}=6\pi m\sqrt{3}
\end{equation}

I hope I got the point of your question, at least partially.