Why are parallelizable manifolds called parallelizable? Let $M$ be a smooth manifold of dimension $n$. Then $M$ is called parallelizable if there exists a global frame, i.e., $n$ linearly independent smooth vector fields $X_1,X_2, \dots, X_n$. 
Why is a paralleizable manifold called "parallelizable"? 
I guess it's related to the parallel transport on $M$. We can always choose a Riemannian metric $g$ on $M$ with the corresponding Levi-Civita connection $\nabla$ on the tangent bundle $TM$. Then we can define the parallel transport of a vector $v \in T_pM$ at any point $p \in M$ along any path $C$ starting from $p$. 
Suppose $M$ is connected. Is it true that if $M$ is parallelizable, then we can always obtain a global frame via the parallel transport of a local frame $v_1, \dots, v_n $ of $T_pM$ at any point $p \in M$?
 A: It's not about parallel transport but the ability to reasonably define when two vectors on different tangent spaces are parallel.
The vector fields provide a natural basis on each tangent space and allows us to compare two tangent vectors by comparing their components in this basis.
With this you can decide when two vectors are equal or parallel.
If you are familiar with bundles, parallelizability means that the tangent bundle is globally trivializable.
This is just another way to say that you can describe tangent vectors irrespective of the base point.
In other words, for any two points $x,y\in M$ there is a canonical linear isomorphism $F_{x,y}:T_xM\to T_yM$ which maps each $X_i(x)$ to $X_i(y)$.
(This property defines $F_{x,y}$ uniquely.)
A connection on the tangent bundle gives a way to identify tangent spaces along a curve, but not between any two points.
Parallel transport is defined along curves, not between points.
Parallelizability in the sense of this question has nothing to do with connections.
The global trivialization of the tangent bundle gives you a way to naturally translate tangent vectors between base points, but it is not parallel transport in the usual sense.
The definition of parallelizability involves no metric or connection, so parallel transport doesn't even have enough structure to work in.
You can always get a metric by defining your frame to be orthonormal.
This gives a "global inner product", but the vectors are not generally parallel transported with respect to this metric.
The term "parallel" in the context of the question's definition must be understood in the sense of parallel vectors in a general vector space, not in the sense of parallel transport along curves.
A: I think it's simpler than that -- you can define parallel transport of the vector 
$$
v = a_1 X_1(p) + \ldots a_n X_n(p)
$$
in $T_p M$ to be
$$
v' = a_1 X_1(q) + \ldots a_n X_n(q)
$$
in $T_q M$. These "parallel translates" of $v$ would, if everything were just $\Bbb R^n$, with $n$ standard-unit-vector vector fields, be actually parallel vectors.
As for your last question: I suspect the answer is "no" -- all you have to do is to pick a metric that's a little wonky. But I don't have the heart to write an actual example right now. 
