As picture below, $u\in C^\infty(M,N)$, $(M,g)$ and $(N,h)$ are two smooth Riemannian manifold.
I don't know what mean the $\frac{\partial }{\partial y^1} \circ u$ , it is $\frac{\partial u}{\partial y^1}$ ?
Besides, how to induce a tangent bundle $u^{-1}TN$ from $TN$ make $u^{-1}TN\subset TM$ ? There is not any condition make sure $dim M\ge dim N$.
In fact ,I think I don't know what $u^{-1}TN$ is.
Picture below is from the page 6 of HARMONIC MAPPINGS BETWEEN RIEMANNIAN MANIFOLDS by Anand Arvind Joshi
Unforunately it is not true that $u^{-1}TN$ is a subbundle of $TM$. By definition, let $f : X \to Y$ and $\pi: E \to Y$ be a vector bundle over $Y$. Then the pullback bundle $f^{-1}E$ on $X$ is given by
$$f^{-1}E = \{ (x, v) \in X \times E : f(x) = \pi(v)\}.$$
Roughly speaking, it is a vector bundle over $X$, such that for each $x\in X$ the fiber is $E_{f(x)}$.
To write down a local basis for $f^{-1} E$, let $e_1 (z), \cdots, e_m(z)$ be a local basis of $E$ around $y = f(x)$. Then
$$\{ e_1 (f(x)), \cdots e_k(f(x))\}$$
serve as a local basis of $f^{-1}E$. Note $e_i (f(x) = (e_i \circ f ) (x)$.
The pullback bundle is involved in harmonic map theory because the differential of a mapping $u : M \to N$ is natually a section in the bundle $T^*M \otimes u^{-1} TN$ (it is just saying that $df_x $ is a mapping from $T_xM$ to $T_{u(x)}N$)