Is there a more straightforward way to define the jet bundle? Everywhere I have looked, the jet bundle is defined as the fiber bundle of equivalence classes for the partial derivatives of functions from one manifold to another. However, it is easy to see that the 1-jet bundle for real functions on a manifold $M$ is actually just $\Bbb R \times T ^*M$, and a function $f \in C^\infty(M)$ will have its 1-jet given by $(f(p), \text d f_p)$. If $f : M \to N$ were instead a smooth function between manifolds, the differential is now a linear operator from $TM \to TN$, so at every point $p \in M$, we could identify it with a tensor in $T_p M \otimes T_{f(p)} ^* N$. Thus, the 1-jet bundle could be given by $N \times (TM \boxtimes TN)$ where $\boxtimes$ is the external tensor product bundle. More specifically, every element in $TM \boxtimes TN$ has the form
$$
(p, q, L) \quad (p, q) \in M \times N, L \in T_p M \otimes T_q N,
$$
so the 1-jet of $f$ could be written as $(f(p), (p, f(p), \text d f_p))$. Now I have two questions: is this construction correct? If it is, how could we generalize it to k-jet bundles?
 A: In principle, the constructions you suggest are correct, some of the details are not as far as I can see. For $J^1(M,\mathbb R)$, $T^*M\times\mathbb R$ looks correct. For $J^1(M,N)$ you should have $T^*M\boxtimes TN$. This is a bundle of $M\times N$, the components of the projection are the source and target projections of the jet and the fiber over $(p,q)$ is $L(T_pM,T_qN)$ which is exactly what you expect. But already here you can see that putting $N=\mathbb R$, you have to make some identifications to recover what you propose as a definition of $J^1(M,\mathbb R)$. It is not clear to me, why you would call such a description more straightforward. Of course, you can identify many jet bundles and jet space with more well known objects but probably the notion of jets is very flexible and universal.
I think it is important to realize in the beginning that there is a natural and universal notions of two maps having $k$th order contact in a point and this is the basic equivalence relation used in defining jets. This is often phrased in the way you state in the question (i.e. via Taylor developments in local coordinates), but there are other nice ways to phrase it, which are independent of coordinates. To describe $k$th order contact for functions $f,g:M\to N$ in $p\in M$, you can for example look at smooth curves $c:I\to M$ with $c(0)=p$ and smooth function $\phi:N\to\mathbb R$ an require that for all such choices you get $(\phi\circ f\circ c)^{(i)}(0)=(\phi\circ g\circ c)^{(i)}(0)$ for all $i=0,\dots k$. Alternatively, you can also require that $f(p)=g(p)$, $T_pf=T_pg$ and the iterated tangent maps $TTf$ and $TTg$ and so on for up to $k$ copies of $T$ agree in the fibers over $p$. Once this definition is there, you get a universal description of jets of all orders of functions between manifolds and sections of bundles, including functorial properties, etc. Say for a smooth function $h:N\to P$, and any $k$, you get a map $h_*:J^k(M,N)\to J^k(M,P)$ induced by $f\mapsto h\circ f$ and a map $h^*:J^k(P,M)\to J^k(N,M)$ induced by $g\mapsto f\circ g$. If you try to give explicit descriptions of jet bundles in the spirit of your question, then you'll need an ad hoc description of these induced maps in each case. And going to $k>1$, you would need iterated tangent bundles and their duals, which are not easy to handle either.
