> 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}
and integrating from $0$ to $2\pi$ we obtain $\Delta t=6\pi m\sqrt{3}$.

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^2_{r=3m}=\frac{1}{3}dt^2 \implies \Delta\tau_{r=3m}=\frac{1}{\sqrt{3}}\Delta t
\end{equation}
obtained by setting $dr=d\phi=d\theta=0$ (as the observer is at rest) and $r=3m$. Substituting $\Delta t=6\pi m\sqrt{3}$ in the equation we get  $\Delta\tau_{r=3m}=6\pi m$. What would a distant observer at rest say about an event "taking" $\Delta t=6\pi m\sqrt{3}$ coordinate time? Again, from the Schwartzschild metric
\begin{equation}
d\tau^2_{r=\infty}=dt^2\implies \Delta\tau_{r=\infty}=\Delta t=6\pi m\sqrt{3}
\end{equation}

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

**Edit:**

Previously I answered to the general question regarding the difference between $dt$ and $d\tau$, without focusing on the specific problem, because it seemed to me that the main source of the problem was the "nature" of these two quantities. Now I'll try to focus on the specific problem you posed, which may seem easy but it's a bit subtle.

Let's ask ourselves what's happening here. We are measuring the period of a light wave, that is to say the time difference between two wave crests, at the point $P_1$. Then the wave travels, and it reaches the point $P_2$, where we measure the period as well. You can see that, even if it seems that we are analyzing the same situation, we are no longer dealing with the same space-time interval between the same events. In the first case we are measuring something specific happening at the point $P_1$, while in the second case we are studying something else (even if it may seem identical) happening at the point $P_2$. But what happens in between? The more specific question should be: if I have a wave the **coordinate-time** difference between two wave crests stays the same no matter "where/when" I measure it, no matter the point $(t,r,\theta, \phi)$ my observer at rest is in space-time?

The answer is: yes, if your space-time is stationary.

Indeed, let's focus on the following space-time interval: the difference between the departure of a wave-crest $A$ from point $P_1$ and the departure of a wave-crest $B$ from the same point. In terms of **coordinates** (we're not interested in physical measurements now, we want to see what's happening in our geometrical manifold), this infinitesimal interval will be characterized by $dt_1, dr_1, d\theta_1, d\phi_1$, the last three being zero as we are dealing with two events (departure of crest $A$, departure of crest $B$) happening at the same point $P_1$.

When the crest $A$ reaches the point $P_2$, followed by the crest $B$, it's this interval we are analyzing: the arrival of the crest $A$ in $P_2$, followed by crest $B$ at the same point. This infinitesimal interval will be characterized by $dt_2, dr_2, d\theta_2, d\phi_2$, the last three being zero. How can we be sure that $dt_2=dt_1$? Well, if the **coordinate-time** $\Delta t_{A,P_1\to P_2}$ taken by the crest $A$ to travel from point $P_1$ to point $P_2$ is equal to $\Delta t_{B,P_1\to P_2}$, **coordinate-time** taken by the crest $B$ to travel from point $P_1$ to point $P_2$, then obviously $dt_1=dt_2$.

It happens that $\Delta t_{A,P_1\to P_2}=\Delta t_{B,P_1\to P_2}$ if we are dealing with a stationary space-time. To show this it suffices to see that the integral
\begin{equation}
\Delta t_{P_1 \to P_2}= \int dt
\end{equation}
can be recast as an integral over the path going from $P_1$ to $P_2$ if we re-express $dt$ using the relation
\begin{equation}
g_{\mu\nu}\frac{dx^\mu}{d \lambda}\frac{dx^\nu}{d \lambda}=0
\end{equation}
as
\begin{equation}
dt=\frac{1}{g_{00}}\sqrt{-g_{i0}dx^i-[g_{i0}g_{j0}-g_{ij}g_{00}]dx^idx^j}
\end{equation}
If the spacetime is stationary, there is no dependence on the time-coordinate, but only on the spatial ones. We thus get $\Delta t_{A,P_1\to P_2}=\Delta t_{B,P_1\to P_2}$ and we can say that $dt_1=dt_2=dt$.

I want to stress out that this is really a subtle issue. In general, in GR problem you encounter at the beginning, you always deal with analyzing the same space-time interval, as the photon orbiting, or the infalling particle, etc. In other cases, such as this one, you deal with light waves and you always have a stationary metric, so you can safely set $dt_1=dt_2=dt$.

I didn't want to add confusion but I remember that when I was studying GR for the first time I completely missed it and it puzzled me seeing, in the case of the redshift you just mentioned, sometimes some author say "so let's set $dt_1=dt_2$" because to me it didn't make any sense.