Why smooth section of vector bundle $F\to M$ is $\Gamma(TM) \times \Gamma(TM) \to \Gamma(NM)$ Let $M\subset \tilde{M}$ be the embedded Riemann submanifold,We can construct the vector bundle  $F\to M$ where each fiber is bilinear map $T_pM\times T_pM \to N_pM$,which is a smooth vector bundle.
The question is why the smooth section of this bundle is $\Gamma(TM) \times \Gamma(TM) \to \Gamma(NM)$ where $NM$ is normal bundle of $M$ ,and $\Gamma$ means smooth section of the corresponding vector bundle.
Let's make the theorem more clear,we have the following charaterization lemma:

Let $B$ be a rough section of vector bundle $F$ then we can define the map
$B(X,Y)(p) = B_p(X_p,Y_p)\in N_pM$
then $B$ is a smooth section of $F$ if and only if $B(X,Y)(p)$ is a
smooth section of $NM$ for each smooth vector field $X,Y$


Pick the local trivialization as follows:take the smooth adapted local frame on $U$,then we have $(E_1,E_2,...,E_k)$ as basis for $T_pM$ and $(E'_{k+1},...,E'_n) $ as basis for $N_pM$.
Hence locally on $U$, the section is $B:U\to \pi^{-1}(U)$ is smooth if and only if $U\to \pi^{-1}(U) \to U\times \Bbb{R}^{k}$ is smooth.
The map we want to study the smoothness under the choice of local trivilization associated with the local frame is : $$U\to \pi^{-1}(U) \to U\times \Bbb{R}^{k}\\ p\mapsto (p,(B^l_{i,j}(p)))$$
Hence $B$ is smooth section if and only if $B^{l}_{i,j}(p)$ is smooth in $p$.
If $B_p(X,Y)$  gives a smooth section of $NM$ for any smooth vector field $X,Y$. in particular for $(E_i,E_j)$ ,due to $B_p(E_i,E_j) = B_{i,j}^l(p)E'_l$ ,we have $B^{l}_{i,j}(p)$ is smooth in $p$.
Conversely if all $B^{l}_{i,j}(p)$ is smooth in $p$. due to the arguement above $B_p(E_i,E_j) $ is smooth section for $NM$ for each $i,j\le k$. since any smooth vector field $X,Y$ can be represented as smooth linear combination of $E_i$( that is $X = X^i(p)E_i$ with $X^i(p)$ smooth),then smoothness of $B_p(E_i,E_j)$ implies smoothness of $B_p(X,Y)$.
