Sometimes the same word "mass" is used with different meanings. There are two different quantities associated with the word "mass":
A quantity that physicists usually call "mass", which is an intrinsic property of the object and does not depend on how fast it is moving. I'll use the symbol $m$ for this quantity.
A synonym for the object's energy $E$, but expressed in mass-like units as $E/c^2$. This is sometimes called the object's "relativistic mass" and it does depend on how fast the object is moving (because the object's energy does). I'll use the symbol $m_R$ for this quantity.
We are already familiar with the fact that the kinetic energy of an object is higher when the object is moving faster. "Relativistic mass" is just a synonym for the object's total energy, expressed in mass-like units. From this perspective, consider the question again:
Why does the (relativistic) mass of an object increase when its speed approaches that of light?
Answer: Because the object's energy increases. "Relativistic mass" is just a synonym for the object's energy, expressed in mass-like units. Why did people ever start using the name "relativistic mass" for the object's energy? I don't know. In my experience, most physicists just call it energy.
Here are some equations to help clarify things:
The energy $E$, momentum $p$, speed $v$, and mass $m$ of an object are related to each other according to these equations:
$$
E^2 - (pc)^2 = (mc^2)^2
\hskip2cm
v = \frac{pc^2}{E}
$$
where $c$ is the speed of light. The $m$ in the first equation is what physicists usually mean when they use the word "mass". It is an intrinsic property of the object and does not depend on the object's speed. The object's energy $E$ and momentum $p$ do depend on the speed, and they do so in such a way that the combination $E^2-(pc)^2$ does not depend on the speed. That's why this particular combination is interesting, and that's why the $m$ on the right-hand side of the equation deserves a special name: mass.
To relate this to the "relativistic mass" $m_R$ (which, again, is not used by the majority of physicists in my experience), re-arrange the second equation shown above to get
$$
p = \frac{E}{c^2}v.
$$
If we use $m_R$ as an abbreviation for $E/c^2$, then this becomes
$$
p = m_R v,
$$
which looks superficially like the more familiar low-speed approximation $p=mv$. This resemblance is also misleading, though, because the energy $E$ (and therefore $m_R$) is a function of $v$. The momentum $p$ is not really proportional to the velocity $v$, except approximately when $v\ll c$.