Relating Temporal Coherence and Intensity of Interference Pattern In trying to understand the phenomenon of coherence a bit deeper, I have come to face the following question.
Suppose one uses an interferometer (Micheloson-Morley, Mach-Zehnder, etc) to measure the temporal coherence of a wave. As the wave works its way through the device, it gets split into two parts such that one part travels a slightly longer path and becomes temporally delayed. Then, the two parts are superposed onto each other and the picture is sent to the detector.
It is here where my question comes in. Looking at the detector, how does one determine whether the signal is highly coherent or whether it is not?
I understand that if the signal is monochromatic or very close to being one, then the interference pattern would remain constant in amplitude and retain periodicity. On the other hand, if the signal's spectrum is composed of multiple frequencies, then the interference patters would "live and breathe" in space. 
How do we relate this intensity to the amount of coherence?
 A: Experimental Observation
The first thing to do to understand this phenomenon is look at the Newton's rings from white light, as I discuss in more detail here.
We can only see a narrow band of wavelengths, so we are looking at the sum of the interference patterns with nulls at radiusses $0,\,\sqrt{4\,R\,\lambda},\,\sqrt{8\,R\,\lambda},\,\cdots$ for $\lambda$ varying between $400{\rm nm}$ and $750{\rm nm}$ where $\lambda$ is the wavelength for the light component in question and $R$ the radius of curvature of the spherical surface pressed against the optical flat in the Newton's rings experiment. This means that the first null happens at a range of radiusses that varies only over a range of about $\pm 20\%$ - the "smear"width is well less than the distance between the first and second null. So, even with the spread over visible wavelengths, the first nulls line up pretty well. The second null less well and so forth. You see a series of colored nulls - the coloring is because different wavelengths have their nulls at different positions, but the nulls are still well enough aligned to see their structure. As you move further from the center, the nulls become more tightly packed and the accuracy of the alignment for all visible wavelengths becomes coarser than the null spacing, which means that we can no longer see fringes. This is exactly what happens in Newton's rings - the fringe visibility fades swiftly with increasing distance from the center.
Let's Get Quantitative and Dirty
Suppose we have two beams interfering in an interfeometer; let there be some offset, large (compared with the effect on the instrument) path difference $\Delta$ between them and a measurand $X$. Then the power of the interference in the presence of purely monochromatic light is:
\begin{equation}
\label{BasicInterference}
p(\Delta + X)=\left|\frac{1}{\sqrt{2}}\left(e^{i\,\frac{k}{2}\,(\Delta + X)} - e^{-i\,\frac{k}{2}\,(\Delta + X)}\right) \right|^2 = 2\,\sin\left(\frac{k}{2}\,(\Delta + X)\right)^2 = 1-\cos\left(k\,(\Delta+X)\right)
\end{equation}
where $k$ is the wavenumber. Now suppose there is a spread of wavelengths present, so that we must form the incoherent sum of all these interference terms:
\begin{equation}
p(\Delta + X)=\frac{\int_{-\infty}^\infty\,\left(1-\cos\left(\left(k_0+\frac{\omega}{c}\right)\,(\Delta+X)\right)\right)\,S(\omega)\,\mathrm{d}\omega}{\int_{-\infty}^\infty\,S(\omega)\,\mathrm{d}\omega} = 1- \frac{\int_{-\infty}^\infty\,\cos\left(\left(k_0+\frac{\omega}{c}\right)\,(\Delta+X)\right)\,S(\omega)\,\mathrm{d}\omega}{\int_{-\infty}^\infty\,S(\omega)\,\mathrm{d}\omega}
\end{equation}
where $S(\omega)$ if the spectrum of the spread, $k_0$ is the center wavenumber and $\omega$ the angular frequency deviation from this central wavenumber / frequency. It is now a simple matter to calculate the fringe visibility given the spectrum $S(\omega)$. 
Lorentzian Lineshapes
With $S(\omega) = \left(1+\frac{4\,\omega^2}{\Omega^2}\right)^{-1}$ where $\Omega$ is the full width, half maximum frequency spread, we get:
\begin{equation}
p(\Delta + X)=1-\cos\left(k_0\,(X+\Delta)\right) \,e^{-\left|\Omega\,\frac{X+\Delta}{2\,c}\right|}
\end{equation}
and so the fringe visibility $\mathscr{V}$ is  $e^{-\left|\Omega\,\frac{X+\Delta}{2\,c}\right|}$. As the spectral width increases, the sum of the two beams is almost constant, with a small sinusoidal variation on top. So we see fringes, but they are a feint striation on an almost constant illumination.
If we convert this expression to one in terms of the Full Width, Half Maximum wavelength spread $\Lambda$ we get:
\begin{equation}
\mathscr{V} \approx e^{-\left|\Omega\,\frac{\Delta}{2\,c}\right|} = \exp\left(-\left|\pi \frac{\Lambda\,\Delta}{\lambda_0^2}\right|\right)
\end{equation}
where $\lambda_0$ is the center wavelength. The coherence length is defined as the path difference needed to lower the visibility to $1/e$; this is:
\begin{equation}
L_c=\frac{\lambda_0^2}{\pi\,\Lambda} = \frac{2\,c}{\Omega} = \frac{c}{\pi\,\Delta\nu}
\end{equation}
where $\Delta\nu$ is the (nonangular), full width half maximum frequency spread. We can also motivate the coherence length as the $1/e$ correlation length when we think of the light wave as a random process with power spectral density $S(\omega) =\frac{2\,\sqrt{2}\,\sigma^2}{\sqrt{\pi}\,\Omega}\, \left(1+\frac{4\,\omega^2}{\Omega^2}\right)^{-1}$, where $\sigma$ is the variance. By the Wiener-Khinchin theorem, the autocorrelation function of the light timeseries is $R(\tau) = \sigma^2\,e^{-\left|\tau\frac{\Omega}{2}\right|}$, whence the $1/e$ correlation time is $2/\Omega$ and so the $1/e$ correlation length is $2\,c/\Omega$ as before.
Indeed, it is not hard to show the general result, valid for any spectrum, that the fringe visibility equals the autocorrelation function $R$ of the normalized (unit variance) light timeseries evaluated at the time $\Delta/c$ delay wrought by the path difference $\Delta$:
$$\mathscr{V} = \frac{R\left(\frac{\Delta}{c}\right)}{R\left(0\right)}$$
You prove this result with the Wiener-Khinchin theorem.
Gaussian Lineshapes
If the spectrum is Gaussian, so that $S(\omega) = \exp\left(-\frac{\omega^2}{2\,\Omega^2}\right)$ so that now $\Omega$ is the rms frequency spread, or frequency standard deviation, we get:
\begin{equation}
\mathscr{V} \approx e^{-\frac{\Delta^2\,\Omega^2}{2\,c^2}} = \exp\left(-2\,\pi^2\,\frac{\Delta^2\,\Lambda^2}{\lambda_0^4}\right)
\end{equation}
where now $\Lambda$ is the rms wavelength spread and the $1/e$ coherence length is:
\begin{equation}
L_c=\frac{\lambda_0^2}{\sqrt{2}\,\pi\,\Lambda} = \frac{\sqrt{2}\,c}{\Omega} = \frac{c}{\sqrt{2}\,\pi\,\Delta\nu}
\end{equation}
where $\Delta\nu$ is the rms frequency spread.
A: The short, intuitive answer is that the temporal coherence is related to the fringe visibility as a function of time delay.  If you can delay the one beam a lot (compared to the oscillation time scale) and the fringes are still visible, then you have strong coherence. If, on the other hand, the fringes disappear quickly as you delay the beam, there is little coherence.
Your thought experiment of putting a couple of coherent beams of different colors into an interferometer is a little misleading. This is because the fringe visibility will remain even with long delays, although as you mentioned the interference pattern will include beating between the frequencies.  A better counterexample to a monochromatic laser would be a broadband lamp (as in an FTIR spectrometer). The broadband lamp has a large interference fringe at zero time delay, but since it is incoherent, the fringes disappear almost instantly with any delay. 
