I am [still] confused about gradients, vectors, deformation gradient, etc.

Dear Ajit:  Thank you for this post.  I eagerly await illumination by experts on geometry.  Here is a try on your Question 1.  I wrote some notes on linear algebra.  Vector space is defined on page 4.  I just follow the standard usage of this phrase "vector space" in linear algebra.

Dear Ajit: If you really want to go deep, then please check

http://en.wikipedia.org/wiki/Discrete_exterior_calculus

and the references. 

Yes it is, if placed in the appropriate vector space (and hence it has components, like any other vector). The essence here is the duality pairing between a vector space and its dual. Thus:

A vector space is an additive group whose elements can be (conveniently) multiplied by numbers (e.g., real, complex). The archetype is the familiar set of "tied" vectors of the 3D space. When talking about derivatives and, in general, tensors/operators, there's another important example: given a vector space V, its
dual V' is the set of all linear functionals f:V->R (for real spaces). V' too is a vector space. The important fact to be mentioned here is that a basis in V induces a basis in V'; and that a change of basis in V induces a change of basis in V' with a different coordinate/component transformation rule: This is the basis for "contravariant" versus "covariant". If V is actually Euclidean (thus equipped with the usual scalar product, denoted here by <.,.>), this distinction disappears, because the action of every linear functional can be represented as a dot product: there exists and is unique a vector G in V such that f(v) = <G,v> for every v in V; V and V' have become one and the same (via the identification G->f).

A vector field is a function assigning a vector to each element of a set. An illustrative example: to each element "x" of a subset of a manifold we assign an element of the tangent space at "x". Down to Earth: the wind velocity over Europe is a vector field (defined over a subset of a sphere).

Finally, before "gradient", one needs to talk about the derivative of a function; function which may be defined over quite abstract objects, e.g., manifolds. However, it is sufficient to consider the case of a function defined over some open subset of a normed space (a vector space equipped with a norm |.|, which, for the purpose of taking limits, should be complete), for then everything extends easily (by composing with coordinate maps - charts). Then we have the classic: given E and F normed spaces and an open subset M of E, a function f:M->F is differentiable at x in M iff there exists a linear operator Df(x):E->F such that f(y)=f(x)+Df(x):(y-x)+o(|y-x|^2). When F is R, the field of real numbers, we get a functional, or a "scalar" function/field. Thus Df is an element of E' (the dual of E). This is the nabla operator. Now if the normed space is actually a Euclidean space (the norm being generated by a scalar product <.,.>), then any linear functional can be uniquely represented as Df:v = <G,v>, for any v in E. We call G the gradient of f; it is an element of E, thus a vector (recall the identification of V and V' at point 1.). When F is R^n, the gradient becomes a tensor (still a "vector", but with more indices).

To conclude, when an ambient scalar product is present, we can safely identify the derivative of a function with its gradient. In this context the classical terminology, "deformation gradient", makes perfect sense and is very convenient ("The derivative of the configuration of a body is called the deformation gradient", Marsden and Hughes, Mathematical foundations of elasticity, Dover ed, pp 47).