Why do opposite dipoles of same charge and equal magnitude cancel each other out? 
In carbon dioxide, there are 2 dipoles of equal magnitude and charge pointing in opposite directions. 
Why can't both oxygen atoms obtain a partially negative charge while the carbon atom contains a partially positive charge twice in magnitude? Shouldn't two oxygen atoms pulling on a single carbon atom result in more positivity? 
Why is carbon dioxide truly nonpolar? Why does this make sense?
 A: Sean's answer is definitive, but I think it's worth adding a few comments.

Why can't both oxygen atoms obtain a partially negative charge while the carbon atom contains a partially positive charge twice in magnitude? Shouldn't two oxygen atoms pulling on a single carbon atom result in more positivity?

The oxygen atoms do get a partial negative charge and the carbon atom does get a partial positive charge. The partial positive charge on the carbon is twice the magnitude of the charges on the oxygen atoms - your diagram is a bit misleading in this respect. I would draw it as:

However this does not produce a dipole moment for the reasons Sean explains. The first non-zero moment is the quadrupole moment. Your title asks:

Why do opposite dipoles of same charge and equal magnitude cancel each other out?

and the answer is simply that the dipole moment is a vector, and the dipoles of the two halves of the molecule add in the usual way vectors add. Since the two dipoles are equal and opposite they add up to zero.

Why is carbon dioxide truly nonpolar? Why does this make sense?

Be careful with the word nonpolar. It does not mean has no dipole moment. Actually I'm not sure that polar has a precise meaning but it generally means interacts with electric fields. Carbon dioxide interacts with electric fields because it has a quadrupole moment, and also because it is polarisable. 
A: What's going on here is you're digging into something called the multipole expansion. The first order of the expansion is known as the "monopole" and is just a measurement of the net amount of charge. Add up the total charge in the carbon dioxide, carbon monoxide, and water molecules and you'll get zero. So we say they have no "monopole moment". In equations, we calculate it with either of these formulae:
\begin{align}
Q & = \sum_{i=1}^N q_i \\
 & = \int \rho(\vec{r}) \operatorname{d}^3r.
\end{align}
Next is the "dipole moment". To calculate this, we add up the positions of the charges, as vectors, weighting each position by the amount of charge there. Notice that position can depend on where you call position zero. The situations where it does are all times when the monopole moment is not zero. In other words, as long as there is no net charge, everyone will agree on the dipole moment. In the case of carbon monoxide the electron is more tightly bound to the oxygen atom, so the oxygen end of the atom is more negatively charged, giving carbon monoxide a net dipole moment. Dipole moments are computed using either of these formulae:
\begin{align}
 \vec{p} & = \sum_{i=n}^N q_i \vec{r}_i q_i \\
  & = \int \vec{r}\, \rho(\vec{r}) \operatorname{d}^3 r.
\end{align}
After the dipole moment is the "quadrupole moment". This one is not quite as easy to explain or visualize, but we can apply a pattern. The most basic dipole that everyone agrees on is to take two charges of equal magnitude, but opposite sign, and separate them. You can picture the dipole as an arrow that points from the negative charge to the positive one.
To construct a basic quadrupole, take two dipoles of equal magnitude (length of arrow) but opposite direction (just flip all of the positive and negative charges) and offset them from each-other without rotating the arrows, and you get a quadrupole. In the case of carbon dioxide, you have two dipoles, one pointing from the carbon to an oxygen, and the other pointing to the other oxygen. Because $\mathrm{CO}_2$ is a linear molecule, the net dipole moment cancels out because the total moment is the vector sum of the moments that comprise the system, leaving just the quadrupole moment.
Contrast that with a water molecule. An $\mathrm{OH}$ molecule also has a dipole moment. When you throw in another $\mathrm{H}$ atom to form $\mathrm{H}_2\mathrm{O}$, it doesn't bond in a straight line, but makes an obtuse angle. So the net dipole moment of the $\mathrm{H}_2\mathrm{O}$ is not zero, but it's a fraction, in magnitude, of the dipole formed by half of the charge on the oxygen atom and one of the hydrogen atoms. It is the perfect cancellation in the linear carbon dioxide that make it non-polar.
In equations, we calculate quadrupole moments using the formulae:
\begin{align}
    Q & = \sum_{k=1}^N \vec{r}_k \otimes \vec{r}_k q_k \\
    & = \int \vec{r} \otimes \vec{r} \rho(\vec{r}) \operatorname{d}^3 r,
\end{align}
where $\otimes$ is the tensor product, or outer product, where a matrix is constructed using the components of the position vector $\vec{r}$. So:
$$\vec{r}\otimes \vec{r} = \left[\begin{array}{ccc}
x^2 & xy & xz \\
xy & y^2 & yz \\
xz & yz & z^2
\end{array}\right].$$
Like the dipole moment, the only way for the quadrupole moment to be independent of choice of origin is for the net monopole and dipole moments to be zero. 
