Notice the definition of center of mass $x_c$ for two particles rearranges to
$$x_c = \Big(\frac{m_1}{m_1+m_2}\Big)x_1+\Big(\frac{m_2}{m_1+m_2}\Big)x_2$$
Let's call the ratio $(\frac{m_1}{m_1+m_2})$ "$W_1$". It can can be interpreted as the percentage of mass that object 1 contributes to the system, generally $(\frac{mass of object 1}{total mass})$. To convert to units of percentage simply multiply the resulting decimal by 100%. Notice that $W_1 \leq 1$, that is object 1 (any object) cannot contribute more that 100% of the mass of the system.
In general a weighted average (arithmetic mean) $x_{av}$ of $n$ components is
$$x_{av} = w_1x_1 + w_2x_2 + ... + w_nx_n = \sum_{i=1}^{n} w_ix_i$$
where $w_i$ is the weight (in the statistical sense) of component $x_i$. For convenience we often wish for the weight factors to be normalized, that is,
$$\sum_{i=1}^{n} W_i = 1$$
Notice here I wrote $W_i$, the normalized weight factors, instead of $w_i$, which are not necessarily normalized.
For your physics application of this math, the mass $m_i$ of object $i$ corresponds to the "crude" weight factor $w_i$. To normalize weight factors (so that none is greater than 100%), we divide the crude factor by the total weight of all components:
$$W_i = \frac{w_i}{\sum_{i=1}^{n} w_i}$$
or, for us, the object mass by the total mass of the system:
$$W_i = \frac{m_i}{\sum_{i=1}^{n} m_i} = \frac{m_i}{M} $$
(where "$M$" is a common abbreviation for the total system mass that is, $M = \sum_{i=1}^{n} m_i$.) This is why our weight factors for calculating center of mass consist of the mass of the object divided by the total mass of the system. For two objects ($n=2$), the weight factor $W_i$ for object $i$ is of course
$$W_i = \frac{m_i}{\sum_{i=1}^{2} m_i} = \frac{m_i}{m_1+m_2} $$
Thus,
$$x_c = \Big(\frac{m_1}{m_1+m_2}\Big)x_1+\Big(\frac{m_2}{m_1+m_2}\Big)x_2 \\
= \sum_{i=1}^{2} \Big(\frac{m_i}{m_1+m_2}\Big) x_i \\
= \frac{\sum_{i=1}^{2} m_ix_i}{\sum_{i=1}^{2} m_i} \\
= \frac{1}{M} \sum_{i=1}^{2}x_im_i$$
I've written a few forms of $x_c$ above so you can mull over how this expression is constructed. For the general center of mass of a system of $n$ objects, simply replace $2$ with $n$ in the final expression.
Let's explicitly juxtapose our center of mass formula with the weighted average formula to drive the point home:
$$x_{av} = \sum_{i=1}^{n} W_ix_i \\
\\
x_c= \sum_{i=1}^{n} \Big(\frac{m_i}{M}\Big) x_i$$
Thus, the center of mass is the weighted average of position with respect to mass.
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P.S. If the system has a continuous distribution of mass rather than $n$ discrete chunks (as for any real object that is not being approximated as a dimensionless point particle), instead of
$$x_c = \frac{1}{M} \sum_{i=1}^{n}x_im_i$$
We can use
$$x_c = \frac{1}{M} \int xdm$$
which arises when we take the limit as $n \to \infty$. (Note this form is not the most practical to calculate so we usually make the substitution $dm = \rho dV$ where $\rho$ is the density of the object, which may vary with position.)
