Addendum.
After re-considering the form of your question (e.g. the M†M in the denominator --- as opposed for instance to a single operator M, which suffices for projectors) and reconsulting my copy of Nielsen and Chaung, here are some supplementary details not covered by my previous answer. (I'm posting this as a separate answer due to length, and because I feel that this is even less of an 'explanation' than my previous answer.)
Suppose that our only means of measuring a qubit X is indirect: by a 'weak' interaction with an ancilla A, followed by a measurement on A.
We would like to be able to talk about these as being in a sense a way of measuring X. How might we describe such a measurement in terms of X alone? Well: suppose we can easily prepare A in the initial state $|+\rangle \propto |0\rangle + |1\rangle$, and perform a controlled unitary of the following sort, with X as the control and A as the target:
$$ U \;=\; \left[\begin{matrix} \quad1\quad & 0 & 0 & 0 \\ 0 & \quad1\quad & 0 & 0 \\ 0 & 0 & \cos(\tfrac{\pi}{12}) & \sin(\tfrac{\pi}{12}) \\ 0 & 0 & -\sin(\tfrac{\pi}{12}) & \cos(\tfrac{\pi}{12}) \end{matrix}\right] $$
We then measure A in the standard basis (so that A now stores the measurement result). This transforms the state of X as follows:
$\begin{align*} |\psi_0\rangle_X \;=&\; \alpha|0\rangle_X + \beta|1\rangle_X
\\\\\mapsto&\; \alpha |0\rangle_X \otimes \bigl(\tfrac1{\sqrt 2} |0\rangle_A + \tfrac1{\sqrt 2}|1\rangle_A\bigr) \quad+\quad \beta |1\rangle_X \otimes \bigl(\tfrac1{\sqrt 2}|0\rangle_A + \tfrac1{\sqrt 2}|1\rangle_A\bigr)
\\\\\mapsto&\; \alpha |0\rangle_X \otimes \bigl(\tfrac1{\sqrt 2} |0\rangle_A + \tfrac1{\sqrt 2}|1\rangle_A\bigr) \quad+\quad \beta |1\rangle_X \otimes \bigl(\tfrac{\sqrt 3}2|0\rangle_A + \tfrac12|1\rangle_A\bigr)
\\\\=&\; \bigl( \tfrac{\alpha}{\sqrt 2} |0\rangle_X + \tfrac{\sqrt3\beta}{2} |1\rangle_X \bigr) \otimes |0\rangle_A \quad+\quad \bigl( \tfrac{\alpha}{\sqrt 2} |0\rangle_X + \tfrac{\beta}{2} |1\rangle_X \bigr) \otimes |1\rangle_A
\\\\\mapsto&\;
\begin{cases}
|\psi_1\rangle_X \otimes |0\rangle_A \;\;\propto\;\; \bigl(\tfrac{\alpha}{\sqrt 2}|0\rangle_X + \tfrac{\sqrt 3\beta}{2}|1\rangle_X\bigr) \otimes |0\rangle_A & \;\text{for the result 0; or } \\
|\psi_1\rangle_X \otimes |1\rangle_A \;\;\propto\;\; \bigl(\tfrac{\alpha}{\sqrt 2}|0\rangle_X + \tfrac{\beta}{2}|1\rangle_X\bigr) \otimes |1\rangle_A & \;\text{for the result 1.}
\end{cases}
\end{align*}$
In the equations above, note that if the result of the measurement is c, the final state $|\psi_1\rangle$ of X is proportional to $|\psi'_1\rangle = M_c |\psi_0\rangle$, where we define
$$ M_0 \;=\; \tfrac{1}{\sqrt 2} |0\rangle\langle 0| + \tfrac{\sqrt 3}{2} |1\rangle\langle 1|\;,\qquad
M_1 \;=\; \tfrac{1}{\sqrt 2} |0\rangle\langle 0| + \tfrac{1}{2} |1\rangle\langle 1|\;;$$
and we may verify that the probabilities with which we obtain the measurement results are in each case $\langle \psi'_1 | \psi'_1 \rangle \;=\; \langle \psi_0 | M_c^\dagger M_c | \psi_0 \rangle$.
This is very close to describing the transformation of X in the same way that we describe projective measurements. But is this any sort of measurement, meaningfully speaking? Well: if we can do statistics on the results of multiple iterations of this procedure, and if X is initially in the standard basis, we would notice that there is a bias in when we obtain the '0' result: we obtain it more often when X is initially in the state $|1\rangle$.
If we can sample enough times to distinguish whether the measurement results are distributed more like $(\frac12,\frac12)$ or $(\frac34,\frac14)$, we can determine with high probability whether the qubit is initially in the state $|0\rangle$ or the state $|1\rangle$.
The similarity of the probabilities-and-update formulae to those of projective measurement, and the fact that we can use measurement statistics to get information about the state measured, motivates a generalization of the notion of 'measurement' to include procedures such as the one above: we may describe possible measurement outcomes by one, two, or more operators $M_c$ (which are in fact 'Kraus operators', objects associated to CPTP maps), with outcomes described by a slightly generalized Born rule
$$ \Pr\limits_{|\psi_0\rangle}(\text{result}=c) \;=\; \langle \psi_0 | M_c^\dagger M_c | \psi_0 \rangle \;, $$
where $M_c$ is a Kraus operator associated with your measurement, and with an update rule given by
$$ |\psi_1\rangle \;=\; \frac{M_c |\psi_0\rangle}{\sqrt{\langle \psi_0 | M_c^\dagger M_c | \psi_0 \rangle}} \;.$$
In order for the probabilities to be conserved (so that with certainty at least one of the measurement results occurs), we require $\sum_c M_c^\dagger M_c = I$. This is the more general form in your question, described by Nielsen and Chaung. (Again, this looks slightly better when describing states by density operators.)
General remarks.
In general, any time that we introduce an ancilla (or collection of ancillas) A, interact a qubit (or register of several qubits) X unitarily with A, and then perform a projective measurement on A, this gives rise to a sort of measurement of X; the measurement operators can then be described by some collection of positive-semidefinite operators $M_c$ such that $\sum_c M_c^\dagger M_c \;=\; I$ (again so that probability is conserved).
The more general, weaker measurements described here are more closely related to POVMs, which allow you to easily describe measurement probabilities 'abstractly', without an explicit choice of transformations $M_c$, by providing operators $E_c = M_c^\dagger M_c$ and allowing you to use these in the Born rule to compute probabilities. As I alluded to both above and in my previous response, POVMs can be regarded as describing statistically-available information about a system.
Thinking of measurements in terms of Kraus operators (and in terms of a 'measurement result register' A as above) in this way allows you to subsume the notion of measurement into that of a CPTP map, which is an idea that I enjoy. (However, this doesn't really change things from an analytical standpoint, and isn't something you should worry about if you're not yet comfortable with CPTP maps).
