The other answers give good insight. Here's another perspective.

 Since the Levi-Civita connection is the unique metric and torsion-free connection, to motivate its use we need to convince ourselves that both of these properties are desirable.  I'll note that there is sometimes value in considering non-metric connections, but in the question you addressed why using metric connections make sense for studying geometry. So I guess the real issue is to tackle torsion-free-ness. 
 

In order to address this, the first thing to do is try to understand what torsion really is anyway. There is another [question][1] on Mathoverflow about torsion with some great answers, but let me try to draw some pictures. We'll start with the standard picture of the curvature tensor (for a torsion-free connection). (Edit: I got several comments about how to interpret these pictures. I'll discuss this at the end of the answer)

[![enter image description here][2]][2]

The idea is that we have three vectors $X$, $Y$ and $Z$. Starting at a point $p$ in our space, we use our connection to parallel transport $Z$ an infinitesimal amount along a geodesic in the $X$ direction and then along a curve in the $Y$ direction. We then parallel transport $Z$ in an infinitesimal amount in the $Y$ direction and then in the $X$ direction. The curvature measures the difference between these two parallel transports. In the formula, the Lie bracket term is there to make sure that everything is nice and tensorial. 

What changes if the torsion is non-zero?
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##  [![Torsion diagram][3]][3] ##

In this case, if we parallel transport along a geodesic in the $X$ direction and then along a geodesic in the $Y$ direction (see below for how to make this precise), we get a different point from when we parallel transport in the $Y$ direction first then in $X$ direction. When we take the logarithm of the differences of these points, what's left is $\epsilon^2 T(X,Y)$ (modulo an error of $\approx \epsilon^3 R(X,Y)(X+Y)$, as Robert Bryant pointed out ). Dividing by $\epsilon^2$ and letting $\epsilon$ to zero, we find the picture above. Again, in the formula there is a Lie bracket term to make everything nice and tensorial.

So why do we want a torsion free connection?
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In my opinion, torsion is complicated invariant and is somewhat hard to understand. For curvature, there is a very clear picture of what it means for a space to have positive versus negative curvature (infinitesimal planes coming together versus spreading apart). As such, it's possible to formulate all sorts of theorems in terms of curvature assumptions. On the other hand, torsion is this awkward vector that you get when you compute multiple derivatives. It's not really meaningful for it to be "positive" or "negative," and so it doesn't  affect the analysis in predictable ways. As such, life is often a lot easier when it's not around, and is what makes the Levi-Civita connection so useful.


I should add that there are times where considering connections with torsion makes sense. For instance, on a Lie group it is possible to construct a curvature-free connection whose torsion encodes the Lie algebra. This is a very useful connection, but from an analytic perspective, it's not so clear geometrically how the respective torsions of $SO(3)$ versus the Heisenberg group (for instance) give rise to their very different geometries.
 Another example is in non-Kahler complex geometry, where we can study holomorphic, complex, metric connections, which must have non-zero torsion. But again, even though the torsion is present and necessary, it's often hard to really use it in a meaningful way.


## How to interpret the pictures ##

There was a long discussion about how to interpret the pictures, so I should say a few words about what they mean. Thanks to Robert Bryant and Matt F for their helpful suggestions,

 When I first learned about the concept, I found it helpful to use the diagrams as a schematic without worrying about which particular fiber everything is defined in. You can still use the diagram to see that the curvature and torsion are skew-symmetric in $X$ and $Y$ and that the curvature is a (3,1) tensor whereas the torsion is a (2,1) tensor.

To make the picture slightly more rigorous, we either parallel transport in the direction $X$ by a distance $\epsilon X$ or, (as shown in the picture) we make $X$ a tangent vector whose length is $O(\epsilon)$. We do the same thing with $Y$. On the other hand, we assume that the norm of $Z$ is $O(1)$. To obtain the diagram, we rescale the geometry by $\frac{1}{\epsilon^2}$ and let $\epsilon \to 0$. As Robert Bryant noted, for non-zero epsilon, the $XY$-parallelogram in the first picture does not fully close, but the displacement is essentially $R(X,Y)(X+Y)$, which is $O(\epsilon^3)$. When we rescale and take limits, this error vanishes, which is why the parallelogram closes in the picture. The fact that this picture is infinitesimal in $X$ and $Y$ is also the reason why the geodesics are drawn as straight lines. 

If we want to make everything completely rigorous while keeping track of the various tangent spaces and making sure that the final expression lives in in $T_p M $, things get more complicated. However, in order to show that this can be done, here's one way to formalize it (using a suggestion by @RobertBryant).

We define the point $q = \exp_p(\epsilon(X+Y)$ to be the opposite corner of the parallelogram. We parallel transport $Z$ along the geodesic $\exp_p(tX)$ for $t$ between $0$ and $\epsilon$ and then parallel transport along the curve $\exp_p(\epsilon X+ t Y)$ until we reach $q$. This traces out the left path around the parallelogram, but the second part of the curve is *not* a geodesic.

 We then do the same thing except that we transport first in the $Y$ direction and then in the $X$ direction. This gives us two vectors at $q$, and we take their difference to get a vector. To bring this back to $p$, we can parallel transport the result back to our original point using the geodesic from $q$ to $p$ (whose logarithm is $\epsilon(X+Y)$). The vector that we obtain by doing this is $$\epsilon^2 R(X,Y)Z+O(\epsilon^3),$$

As such, when we renormalize by $\epsilon^2$ and let $\epsilon \to 0$, we get the desired expression. I prefer drawing the curvature at $q$, rather than $p$ because it visually shows that I am commuting two covariant derivatives.

Unfortunately, we can't use this exact idea for the second picture, because here it really matters that all of the curves are geodesics with respect to the connection $\nabla$. Instead, we travel along the geodesic $\exp_p^\nabla(tX)$ until we hit the top left corner. Then we travel along a geodesic in the "direction" $Y$ (more precisely, the parallel translate of $Y$ along the geodesic from $p$ to $\exp_p^\nabla(\epsilon X)$. We then do the same thing except that we first travel in the $Y$ direction and then the "$X$ direction" (with the same caveat as before). When we do this, the resulting "parallelogram" doesn't close up, and if we take the logarithm of the differences, what we obtain is 
$$\epsilon^2 T^\nabla(X,Y)+\epsilon^3 R^\nabla(X,Y)(X+Y) + \epsilon^3 T^\nabla(T^\nabla(X,Y),X+Y)+O(\epsilon^4),$$ after we parallel transport the vector from $q$ back to $p$. Normalizing by $\epsilon^2$ and letting $\epsilon \to 0$, we get the torsion exactly.


  [1]: https://mathoverflow.net/questions/20493/what-is-torsion-in-differential-geometry-intuitively
  [2]: https://i.sstatic.net/RSrOW.png
  [3]: https://i.sstatic.net/lZHNh.png