In how many ways can an iterated tangent bundle (T^k)M be viewed as a fibre bundle over (T^(k-1))M?
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.
AMENDED POST: Marco Gualtieri points out a fundamental mistake I made. I retract my assertion that there infinitely many such maps. It looks to me there is a surjective map $T^n M \to T^{k}M$ for each monotone injective map $\lbrace 1,\dots,k\rbrace \to \lbrace 1,\dots, n\rbrace$.
ORIGINAL POST: I believe that "infinitely many" is the answer.
We can think of $TM$ as the effect of applying an equivalence relation (two paths
are equivalent if they have equal one jets at the origin) on the space of smooth paths $\Bbb R\to M$, which I'll denote as $M^{\Bbb R}$. Similarly, the $n$-fold iterated tangent bundle $T^nM$ is given by a similar equivalence relation on $M^{\Bbb R^n}$. For any linear embedding $\Bbb R^k \to \Bbb R^n$ there will be a restriction map $T^nM \to T^kM$.
This gives infinitely many maps.
Similarly, for any linear surjection $\Bbb R^n \to \Bbb R^k$ one gets an embedding $T^kM \to T^nM$.
If we use only coordinate inclusions and coordinate projections, then the structure you get is a simplicial object $X.$ with $X_k = T^{k+1}M$ augmented over $M$. This is because $M \mapsto TM$ is a triple (comonad). I've always felt that one could make sense of the de Rham complex from this point of view (somehow).