Lee's proof that the normal bundle is embedded First I'll fix some of the definitions and context for the forthcoming question.
Suppose $M\subseteq \Bbb R^n$ is an embedded $m$-dimensional submanifold. For each $x\in M$, we define the normal space to $\pmb M$ at $\pmb x$ to be the $(n-m)$-dimensional subspace $N_xM\subseteq T_x\Bbb R^n$ consisting of all vectors that are orthogonal to $T_xM$ with respect to the Euclidean dot product. The normal bundle of $\pmb{M}$, denoted by $NM$, is the subset of $T\Bbb R^n\approx \Bbb R^n\times\Bbb R^n$ consisting of vectors that are normal to $M$:
$$
NM = \big\{(x,v) \in \Bbb R^n\times\Bbb R^n : x\in M,\ v\in N_xM \big\}.
$$
The statement and part of the proof of Theorem 6.23 in Lee's Introduction to Smooth Manifolds is reproduced below:

Theorem 6.23. If $M\subseteq \Bbb R^n$ is an embedded $m$-dimensional submanifold, then $NM$ is an embedded $n$-dimensional submanifold of $T\Bbb R^n\approx \Bbb R^n\times\Bbb R^n$.
Proof. Let $x_0$ be any point of $M$, and let $(U,\varphi)$ be a slice chart for $M$ in $\Bbb R^n$ centered at $x_0$. Write $\widehat U = \varphi(U)\subseteq \Bbb R^n$, and write the coordinate functions of $\varphi$ as $\big(u^1,\dots,u^n\big)$, so that $M\cap U$ is the set where $u^{m+1}=\dotsb=u^n=0$. At each point $x\in U$, the vectors $E_j|_x = (d\varphi_x)^{-1}\big(\partial/\partial u^j|_{\varphi(x)}\big)$ form a basis for $T_x\Bbb R^n$. We can expand each $E_j|_x$ in terms of the standard frame [emphasis added] as
  $$
E_j\big|_x = E_j^i(x)\frac{\partial}{\partial x^i}\bigg|_x,
$$
  where each $E_j^i(x)$ is a partial derivative of $\varphi^{-1}$ evaluated at $\varphi(x)$, and thus is a smooth function of $x$.

The proof goes on, but my question is about this part of the proof.


*

*Is $\partial/\partial u^j|_{\varphi(x)}$ literally the partial derivative operator in the $e_j = (0,\dots,0,\underbrace{1}_{\text{$j$th component}},0,\dots,0)$ direction? If not, what is it precisely, using Lee's notation?

*I suspect that if $\partial/\partial u^j|_{\varphi(x)}$ is not literally the partial derivative operator in the $e_j$ direction, then
$\partial/\partial x^j|_x = (d\varphi_x)^{-1}\big(\partial / \partial x^j|_{\varphi(x)}\big)$, where $\partial / \partial x^j|_{\varphi(x)}$ is
literally the partial derivative operator in the $e_j$ direction, and that $\partial/\partial u^j|_{\varphi(x)}$ must be something else.

*What is $E^i_j(x)$ explicitly?


Note that Lee hasn't yet defined standard frame at this point in the text, so I suppose a complete answer to this question would also address what the standard frame actually is.
 A: Kelvin Lois' answer essentially got me "unstuck," but I want to write up my own answer to this question, since I find that by writing things out in detail, not making any of the "usual" identifications makes it easier for me to understand, and perhaps it may help others too.
Instead of writing the coordinate functions of $\varphi$ as $\big(u^i\big)$, let's write $\varphi = \big(\varphi^i\big)$, and reserve the letters $u^i$ to denote the canonical coordinates on the codomain of $\varphi$. If $p\in U$, then
$$
E_j|_p = \frac{\partial}{\partial u^j}\bigg|_p \stackrel{\text{def}}{=} d\big(\varphi^{-1}\big)_{\varphi(p)}\bigg(\frac{\partial}{\partial u^j}\bigg|_{\varphi(p)}\bigg), \quad j=1,\dots,n,
$$
and the $E_j|_p$ form a basis for $T_p\mathbb{R}^n$. Let us write the coordinate functions of the identity map of $\mathbb{R}^n$ as $I = \big(\pi^i\big)$, and use the letters $x^i$ for the canonical coordinates on the codomain of $I$, considered as a chart $\big(\mathbb{R}^n,I\big)$. Then we get another basis for $T_p\mathbb{R}^n$, the standard basis:
$$
\frac{\partial}{\partial x^i}\bigg|_p \stackrel{\text{def}}{=} d\big(I^{-1}\big)_{I(p)}\bigg(\frac{\partial}{\partial x^i}\bigg|_{I(p)}\bigg),\quad i = 1,\dots,n,
$$
so we can express $E_j|_p$ in terms of the standard basis:
$$
E_j|_p = E_j^i(p)\frac{\partial}{\partial x^i}\bigg|_p.
$$
Now we can follow Lee on page 64 to get
\begin{align*}
\frac{\partial}{\partial u^j}\bigg|_p &= d\big(\varphi^{-1}\big)_{\varphi(p)}\bigg(\frac{\partial}{\partial u^j}\bigg|_{\varphi(p)}\bigg) \\
&= d\big(I^{-1}\big)_{I(p)}\circ d\big(I\circ \varphi^{-1}\big)_{\varphi(p)}\bigg(\frac{\partial}{\partial u^j}\bigg|_{\varphi(p)}\bigg) \\
&= d\big(I^{-1}\big)_{I(p)}\bigg(\frac{\partial \big(\pi^i\circ \varphi^{-1}\big)}{\partial u^j}\big(\varphi(p)\big)\frac{\partial}{\partial x^i}\bigg|_{I(p)}\bigg) \\
&= \frac{\partial \big(\pi^i\circ \varphi^{-1}\big)}{\partial u^j}\big(\varphi(p)\big)\,d\big(I^{-1}\big)_{I(p)}\bigg(\frac{\partial}{\partial x^i}\bigg|_{I(p)}\bigg) \\
&= \frac{\partial \big(\pi^i\circ \varphi^{-1}\big)}{\partial u^j}\big(\varphi(p)\big)\,\frac{\partial}{\partial x^i}\bigg|_{p}.
\end{align*}
Now the map $\varphi\colon U\to \widehat U$ is smooth (it is a diffeomorphism), and so is the partial derivative $\frac{\partial (\pi^i\circ \varphi^{-1})}{\partial u^j}\colon \widehat U\to\mathbb{R}$ of the transition map, so the composition $E_j|_{\cdot}=\frac{\partial (\pi^i\circ \varphi^{-1})}{\partial u^j}\circ\varphi\colon U\to\mathbb{R}$ is also smooth.
A: For your first question, just look at $\Bbb{S}^2 \subset \Bbb{R}^3$. The spherical coordinates $(u^1,u^2,u^3)=(\theta,\phi,\rho)$ form a slice chart for $\Bbb{S}^2$, but we knew that not one of the basis vector $\{\partial_{\theta}, \partial_{\phi},\partial_\rho\}$ is equal to any of  $\{\partial_x,\partial_y,\partial_z\}$. (This is exercise 5.10 actually). 
This is obvious since both are different charts $(U,x^i)$ and $(V,\widetilde{x}^i)$, and we don't expect their basis to be equal, but related by a coordinate transformation rule $\partial_{x^i}|_p = \partial \widetilde{x}^j/\partial x^i (\hat{p}) \, \partial_{\widetilde{x}^j}|_p$.
To obtain explicit form of $E^i_j(x)$, just carry out the computation $E_j|_x (x^k)$, where $x^k : U \to \Bbb{R}$ is the $k$-th coordinate function; so
\begin{align}
E^k_j(x) &= E_j|_x x^k =  (d\varphi_x)^{-1} \Bigg(\frac{\partial}{\partial u^j}\Big|_{\varphi(x)}\Bigg) x^k \\ &= d(\varphi^{-1})_{\varphi(x)} \Bigg(\frac{\partial}{\partial u^j}\Big|_{\varphi(x)} \Bigg) x^k = \frac{\partial}{\partial u^j}\Big|_{\varphi(x)} (x^k \circ \varphi^{-1}) \\&= \frac{\partial \big(\varphi^{-1}\big)^k}{\partial u^j} (\varphi(x)).
\end{align}
Therefore $$E_j|_x = \frac{\partial \big(\varphi^{-1}\big)^i}{\partial u^j} (\varphi(x)) \frac{\partial}{\partial x^i}\Big|_{x}$$
Standard frame is not yet defined until Ch.8 (about vector field). Here standard frame is a coordinate frame of the standard chart for $\Bbb{R}^n$ (identity map).
A: The map $\varphi$ is a local diffeomorphism. Suppose $\varphi(x) =0$,as you guessed, let $e_1,...,e_n$ be a basis of $\mathbb{R}^n$, $\varphi^{-1}(x_1,..,x_m,0,..,0)$ is in $M$, this implies that ${\partial\over\partial_i}\varphi^{-1}(x_1,..,x_m,0,..0)=E_i(x)$ is tangent to $M$. The standard frame is just the canonical basis. If $(x_1,...,x_n)$ are the coordinates $e_i={\partial\over\partial_i}(x_1,...,x_n)$.
