If we use the notation $(TM, p_M, M)$ for the tangent bundle of any manifold $M$, then you are right to think that $T^{\ k}M$ has $k$ natural vector bundle structures over $T^{\ k-1}M$ and so on down to $M$, making a diagram which is a $k$-dimensional cube. Such a structure is a $k$-fold vector bundle (See articles by Kirill Mackenzie) and $T^{\ k} M$ is a particularly symmetrical one. An easy way of writing the $k$ bundle maps would be
$$T^{\ a}(p_{T^{\ b} M}) $$
for $a+b+1 = k$, where this means that we are taking the $a$-th derivative of the tangent bundle projection $T\ (T^{\ b}M)\to T^{\ b}M$, yielding a map now from $T^{\ k}$ to $T^{\ k-1}$.
To see the symmetries of $T^{\ k} M$ it is more convenient to describe the functor in a direct way rather than as a $k$-fold composition -- just as you might think of tangent vectors as infinitesimal curves, you can think of points in $T^{\ k} M$ as infinitesimal maps of a unit $k$-cube into $M$. The restrictions to the $k$ faces of the cube (going through the origin) gives your $k$ maps.
From that point of view, it is clear that you can permute the $k$ coordinate axes and get another map of a cube into $M$, so that the functor $T^{\ k} M$ has a $S_k$ group of natural-automorphisms.
Incidentally, to make the above into a definition of $T^{\ k} $, you could do the following:
Consider the fat point $fp$, which you should think of as a space whose smooth functions form the ring $\mathbb{R}[x]/(x^2)$. Then the tangent bundle $TM$ can be thought of as the space of maps from the fat point to $M$, i.e. $TM=C^\infty(fp,M)$. Such maps, by the way, are just algebra homomorphisms from the algebra $C^\infty(M,R)$ to $\mathbb{R}[x]/(x^2)$. You can check that such a map has two components $f_0 + f_1 x$, and that $f_0$ is a homomorphism to $\mathbb{R}$ defining a maximal ideal (i.e. a point $p$ in $M$) and $f_1$ defines a derivation (i.e. a vector at $p$).
In precisely the same way you can consider a cubical fat $k$-point $kfp$ with functions
$$\mathbb{R}[x_1,...,x_k]/(x_1^2,...,x_k^2)$$
and then define $T^{\ k} M = C^\infty(kfp,M)$. Then you can see the symmetries as automorphisms of the above algebra, and the $k$ maps you ask about as homomorphisms $C^\infty(kfp)\to C^\infty\big((k-1)fp\big).$
There are probably more subtle things to be said about these higher iterated bundles but I hope the above is at least correct.
Here's a sketch of the relation between div-grad-curl and the de Rham complex, in case you might find it useful.
The first thing to realise is that the div-grad-curl story is inextricably linked to calculus in a three-dimensional euclidean space. This is not surprising if you consider that this stuff used to go by the name of "vector calculus" at a time when a physicist's definition of a vector was "a quantity with both magnitude and direction". Hence the inner product is essential part of the baggage as is the three-dimensionality (in the guise of the cross product of vectors).
In three-dimensional euclidean space you have the inner product and the cross product and this allows you to write the de Rham sequence in terms of div, grad and curl as follows:
$$ \matrix{ \Omega^0 & \stackrel{d}{\longrightarrow} & \Omega^1 & \stackrel{d}{\longrightarrow} & \Omega^2 & \stackrel{d}{\longrightarrow} & \Omega^3 \cr
\uparrow & & \uparrow & & \uparrow & & \uparrow \cr
\Omega^0 & \stackrel{\mathrm{grad}}{\longrightarrow} & \mathcal{X} & \stackrel{\mathrm{curl}}{\longrightarrow} & \mathcal{X} & \stackrel{\mathrm{div}}{\longrightarrow} & \Omega^0 \cr}$$
where $\mathcal{X}$ stands for vector fields and the vertical maps are, from left to right, the following isomorphisms:
- the identity: $f \mapsto f$
- the musical isomorphism $X \mapsto \langle X, -\rangle$
- $X \mapsto \omega$, where $\omega(Y,Z) = \langle X, Y \times Z \rangle$
- $f \mapsto f \mathrm{dvol}$, where $\mathrm{dvol}(X,Y,Z) = \langle X, Y \times Z\rangle$
up to perhaps a sign here and there that I'm too lazy to chase.
The beauty of this is that, first of all, the two vector calculus identities $\mathrm{div} \circ \mathrm{curl} = 0$ and $\mathrm{curl} \circ \mathrm{grad} = 0$ are now subsumed simply in $d^2 = 0$, and that whereas div, grad, curl are trapped in three-dimensional euclidean space, the de Rham complex exists in any differentiable manifold without any extra structure. We teach the language of differential forms to our undergraduates in Edinburgh in their third year and this is one way to motivate it.
As for the integral theorems, I always found Spivak's Calculus on manifolds to be a pretty good book.
Another answer mentioned Gravitation by Misner, Thorne and Wheeler. Personally I found their treatment of differential forms very confusing when I was a student. I'm happier with the idea of a dual vector space than I am with the "milk crates" they draw to illustrate differential forms. Wald's book on General Relativity had, to my mind, a much nicer treatment of this subject.
Best Answer
Your $DX$ generalizes to the setup where $X$ is a section $M\to E$ of a vector bundle $E$ on $M$, giving a (fiberwise linear) map $DX:TM\to TE$. At each point $p\in M$ this gives a linear map $D_pX:T_pM\to T_{X(p)}E$. There is a canonical short exact sequence $0\to E_p\to T_{X(p)}E\to T_pM$. The composed map $T_pM\to T_{X(p)}E\to T_pM$ is the identity, so the "TM part" of $DX$ does not depend on $X$.
A connection on $E$ gives a splitting of that exact sequence, so in the presence of a connection you can speak of the "E part" of your total derivative. That's $\nabla X$.
So you can sort of say that $\nabla X$ knows no more and no less than your $DX$ (but it doesn't appear until you choose a connection).
Note that $\nabla_YX$ is linear over the functions: we have $\nabla_{fX}Y=f\nabla_YX$ for a smooth function $f$ on $M$.
The Lie derivative is rather different. It's not linear over the functions in the $Y$ variable, and it's only defined when $E=TM$. Maybe someone else can say something more positive about where it fits in.