Why torsion is only defined for linear connection on TM? The concept of curvature is defined for any linear connection on any vector bundle $E \to M$, but the concept of torsion is only defined for connection on the tangent bundle $TM$ of a manifold $M^n$, or for a connection obtained as the pullback of a connection on a vector bundle $E \to M$ isomorphic to $TM$ via an isomorphism $\theta \colon TM \to E$ equivalent to a solder form.
Why is that so ? If torsion can be interpreted as the twist of a moving frame along a curve, the same phenomena should occur for a connection on any vector bundle.
Is there a way to define a notion of torsion for any vector bundle ?
 A: The torsion of a connection on the tangent bundle is a zeroth order invariant tensor. There is no zeroth order invariant tensor associated to a connection on a principal bundle on a smooth real manifold, since in any coordinates, the connection is $A=g^{-1} \, dg + \operatorname{Ad}_g^{-1} \Gamma_i(x) \, dx^i$, and we can then change those coordinates to get $x=0$ at our chosen point, $g=1$ there, and $\Gamma_i(0)=0$, by replacing $g$ by $h(x)g$ so that we replace $A$ by $g^{-1} \, dg + \operatorname{Ad}_g^{-1} \operatorname{Ad}_{h(x)}^{-1}(\Gamma_i(x) \, dx^i+dh \, h^{-1})$. We pick $h(x)$ so that $dh  \, h^{-1}(0)=-\Gamma_i(0) dx^i$. So all connections look the same at a point, to first order. You can only feel that the connection is not flat at second order. This doesn't work on the frame bundle (the principal bundle associated to the tangent bundle) because you can't change $g$ to $h(x)g$ independent of how you change coordinates $x$.
A: Actually, one can define torsion for Lie algebroids, and in particular vector bundles, eg. here: https://arxiv.org/pdf/math/0105033.pdf
It is defined as so: given a Lie algebroid $A\to M$ with anchor map $\alpha\,,$ choose a connection $\nabla\,.$ The torsion tensor is then defined as $T(X,Y)=\nabla_{\alpha(X)}Y-\nabla_{\alpha(Y)}X-[X,Y]\,.$
When the Lie algebroid is the tangent bundle, you get the usual torsion. When the Lie algebroid is a Lie algebra, you get the commutator (up to a sign). When the Lie algebroid is a vector bundle (by which I mean the anchor map and bracket are zero) the torsion is zero, which is consistent with what the other commentors have said.
A: I believe  that  no ``natural'' section of either $E^*\otimes E^*\otimes E$  or $T^*\otimes T^*\otimes E$ or even $T^*\otimes E^*\otimes E$ can be  associated with a general connection on a vector bundle $E\to M$, although I  would be interested to see a nice proof of this for myself. (I do not recall anything similar in the literature.)  Geometrically, torsion is a measure how far is a connection from being symmetric, but there is no reasonable notion of a symmetric connection for  a general vector bundle. 
