How can we figure out how fast the coffee is warming up the mug? The hot coffee is warming up the cold mug.
How fast is that happening? Well, to answer that question, we have to know how fast this is happening:
The cold mug is cooling down the hot coffee.
But how fast is that happening? Well, to answer that question, we have to know how fast this is happening:
The hot coffee is warming up the cold mug.
And so on, to infinity. The loop seems to me to be infinite. How can we break out of it?
 A: If the hotter object was at $T_H$, the colder one at $T_C$ then heat would flow from the hotter to the colder one (we assume no heat is lost to the environment) and the heat flux $\frac{dQ}{dt}$ is given by:
$$\frac{dQ}{dt}=kA\Delta T$$
where $k$ is the heat transfer coefficient, $A$ the surface area connecting the objects and $\Delta T$:
$$\Delta T=T_H-T_C$$
Assume the heat capacity of the hotter object is $C_HM_H$, of the colder object it is $C_CM_C$, then an infinitesimal loss of heat energy $dQ$ is given by:
$$dQ=C_HM_HdT_H=-C_CM_CdT_C$$
So:
$$dT_H=-\frac{C_CM_C}{C_HM_H}dT_C$$
Set:
$$\alpha =\frac{C_CM_C}{C_HM_H}$$
So:
$$dT_H=-\alpha dT_C$$
We also know from Conservation of Energy that the shared end-temperature $T_E$ is given by:
$$C_HM_HT_H+C_CM_CT_C=(C_HM_H+C_CM_C)T_E$$
So:
$$T_H+\alpha T_C=(1+\alpha)T_E$$
$$T_H=(1+\alpha)T_E-\alpha T_C$$
$$\Delta T=(1+\alpha)T_E-\alpha T_C-T_C=(1+\alpha)(T_E-T_C)$$
So with the above:
$$-\frac{C_CM_CdT_C}{dt}=kA(1+\alpha)(T_E-T_C)$$
$$\frac{dT_C}{T_E-T_C}=-\beta dt$$
where:
$$\beta=\frac{kA(1+\alpha)}{C_CM_C}$$
Integrating between $T_C$ and $T_{C,t}$ we get:
$$\ln \frac{T_E-T_{C,t}}{T_E-T_C}=-\beta t$$
Or:
$$\large{T_{C,t}=T_E+(T_E-T_C)e^{-\beta t}}$$
So $T_{C,t}$ tends asymptotically to $T_E$:
$$t \to \infty\:\text{then} T_{C,t} \to T_E$$
The same is of course true of $T_{H,t}$ and also $t \to \infty$, then $\Delta T \to 0$.
Schematically, temperature evolution of both objects looks like:

A: Dmitry handled the heart of your confusion perfectly in the comments, but I figured I would at least answer your question:

How can we figure out how fast the coffee is warming up the mug?

Physics is an experimental science.  As much as the pop science community glorifies thought experiments about twin astronauts and cats, in this field we answer questions by measurement.

The hot coffee is warming up the cold mug.  How fast is that happening?

We time it.  Put a thermometer on the outside of a room temperature coffee mug, fill it with coffee and start your stopwatch.  Every millisecond, or as frequently as you're legally allowed to push your indentured-servant-graduate-student, write down the time and temperature on a pad of paper.  Plot and publish.

This answer wasn't meant to be cheeky.  I'm sure you knew you could answer your question this way, but you still thought there was a paradox.  In general, whenever your theory has an impossibility in it but your experiment doesn't, you need to change your theory (you aren't the first person to make this mistake).
In this case, as comments pointed out, just define the heat as flowing one way and you're done.
A: Your question was 

The loop seems to me to be infinite. How can we break out of it?".

We get out of it by realizing that when two bodies are in contact, heat will flow from the hotter to the cooler object. In the process, the cool object will get hotter. The rate of heat flow is a function of the thermal conductivity of the material, the thermal gradient, and the heat capacity. And that is a problem that can be solved mathematically - no paradox.
To simplify the analysis, we can look at the case of two identical semi-infinite rods, of different temperature, that are brought in close thermal contact at t=0. If the thermal profile at a given point in time $t$ and space $x$ is $u(x,t)$ then we can analyze the heat flow. This gives rise to the diffusion equation - see for example this lecture for an detailed analysis and explanation. In short, if you have thermal conductivity $K$, specific heat $\sigma$, cross sectional area $A$ and density $\rho$, we can write for the heat flow across a surface at $x$:
$$\Phi(x) = -KA\left.\frac{\partial T}{\partial x}\right|_x$$
At a point $x + \delta x$ we obtain
$$\Phi(x+\delta x) = -KA\left.\frac{\partial T}{\partial x}\right|_{x+\delta x}$$
The difference is the heat that is available to heat up the little slice $\delta x$. Conservation of heat:
$$KA\frac{\partial^2u}{\partial x^2}\delta x = \sigma \rho A \delta x \frac{\partial u}{\partial t}$$
Writing $c^2 = \frac{K}{\sigma \rho}$, this reduces to the diffusion equation:
$$\frac{\partial u}{\partial t} = c^2 \frac{\partial^2u}{\partial x^2}$$
You can use Fourier analysis to solve this for any configuration and boundary conditions. Bottom line is that the heat that is leaving the hotter object will warm up the cooler object, and reduce the thermal gradient. This will slow down the heat flow. 
If you are interested, there is quite an extensive set of cases solved in this paper.
And here is a diagram of how the heat diffusion causes an initial step function to "diffuse" with time (from this lecture):

