How I express the Jacobian matrix between xy-plane and cylinder? When I review the differential two manifolds $F:M\to N$, I find that I do not surely understand this concpt. Let $M$ be $xy$-plane, i.e $ M=\left\{ (x,y,0):x,y \in \mathbb{R} \right\}$ , and $N=\left\{ (x,y,z):x^2+y^2= R^2 \right\}$. I want to express $J(F)$ Jacobian matrix, but I cannot express $J(F)$.  First, Let $v=\sum_{i=1}^{n}a_{i}\frac{\partial}{\partial x_i}|_{p}  \in T_pM , \varphi \in C^\infty{(N)} $ Then, the definition of differential of two manifolds $dF_p(v)$ is defined as
$$dF_p(v)(\varphi):= v(\varphi \circ F)= \sum _{i=1}^{m} a_i \frac{\partial (\varphi \circ F)}{\partial x_i}|_{p} .........(♣) $$
Then, clearly since $dF_p(v)$ is a linear operator(or linear mapping). Then, it is meaningful to express $dF_p$ : $T_p(M) \to T_{F(p)}N, ( v \mapsto dF_p(v))$ as a $2\times 2$ matrix. Then, $F(x,y)=(R \operatorname{cos}(\frac{x}{R}), R \operatorname{sin}(\frac{x}{R}), y)$.
And next, naturally, $T_pM$ is $M$ itself (so I choose a standeard basis $\mathfrak{B}= \left\{(1,0,0)(=\frac{\partial}{\partial x_1}|_{p}), (0,1,0)(=\frac{\partial}{\partial x_2}|_{p})\right\}$) and $T_{F(p)}N$ is generated by $\partial F_x= \left(-\operatorname{sin}\frac{x}{R}, \operatorname{cos}\frac{x}{R},0\right)$, $\partial F_y= \left(0,0,1\right)$. Hence, the basis of $T_{F(p)}$, denoted by $\mathfrak{C}=\left\{\left(-\operatorname{sin}\frac{x}{R}(=\frac{\partial}{\partial y_1}|_{F(p)}), \operatorname{cos}\frac{x}{R},0\right), (0,0,1)(=\frac{\partial}{\partial y_2}|_{F(p)})\right\}$. So I think that, $dF_p = [T]_{\mathfrak{B}}^{\mathfrak{C}}$, or
$$dF_p(e_1)= b_{11} \partial F_x+ b_{12} \partial F_y$$
$$dF_p(e_2)= b_{21} \partial F_x+ b_{22} \partial F_y$$
Next, I think that $dF_p$ can be expressed differently by using the definition (♣). Then, I can find each component of matrix $b_{ij}$. And acutally,
$$J(F)=\begin{bmatrix}
b_{11} & b_{12}\\
b_{21} & b_{22} \\
\end{bmatrix}$$
will be a matrix I want to get, ​But I am now stuck this part.
​From here, according to the textbook I follow , the reader "choose" a function $\varphi : V (\subset N) \to \mathbb{R}$  Then is it possible to choose any( but easier to compute) $c^{\infty}$-function?
According to the textbook,

in order to find Jacobian matrix, if
$v=\sum_{i=1}^{n}a_{i}\frac{\partial}{\partial x_i}|_{p}$
$$df_p(v)=\sum_{i=1}^{n}b_j \frac{\partial}{\partial y_j}|_{f(p)} $$
(The author does not decompose the vector as component as I did...)
our aim is to express $b_j (j=1,2,...m)$ as $a_i(i=1,2,....,n)$
In order to find $b_j$, pick a $\varphi : V (\subset N) \to
 \mathbb{R}$, and apply this function $\varphi$ to (♣), then
$$\left(\sum_{i=1}^{n}b_j \frac{\partial}{\partial y_j}|_{f(p)}\right)
 \varphi =...= \sum _{i=1}^{m} a_i \frac{\partial (\varphi \circ
 F)}{\partial x_i}|_{p}   $$
Also, by chain rule,
$$\sum _{i=1}^{m} a_i \frac{\partial (\varphi \circ F)}{\partial
 x_i}|_{p}= \sum _{i=1}^{m} a_i\sum_{i}^{n} \frac{\partial
 \varphi}{\partial y_j}|_{F(p)} \frac{\partial y \circ F}{\partial
 x_i}_{p}= \sum_{i=1}^{n} \left(\sum_{j=1}^{m} \frac{\partial
 \overset{\sim}{f_j}}{\partial x_i}|_{p} a_{i} \right)
 \frac{\partial}{\partial y_j}|_{F(p)} \varphi $$
(where $\overset{\sim}{f_j}:=y_j \circ f$)
Hence, $$b_j=\sum_{j=1}^{m} \frac{\partial
 \overset{\sim}{f_j}}{\partial x_i}|_{p} a_{i} $$
Thus Jacobian matrix (between two mainfolds), $J(F)=\left\{ \frac{\partial
 \overset{\sim}{f_j}}{\partial x_i}|_{p}\right\}_{1\le i \le n, 1\le j
 \le m}$

The reference I see is a general case, so I apply this general part for specific example, but I am now stuck as I already said, how to choose function $\varphi$.
 A: Let's use coordinates $u,v$ in the domain and $\theta,z$ in the image. As you've said, it's natural to use $\frac{\partial}{\partial u},\frac{\partial}{\partial v}$ as a basis for the tangent space of the domain (at every point) and, similarly, $\frac{\partial}{\partial \theta},\frac{\partial}{\partial z}$ as a basis for the tangent space of the image (at every point).
Note that we parametrize the cylinder $x^2+y^2=R^2$ by $(R\cos\theta,R\sin\theta,z)$, and so $\dfrac{\partial}{\partial\theta} = R(-\sin\theta,\cos\theta,0)$ and $\dfrac{\partial}{\partial z} = (0,0,1)$.
With your mapping $F(u,v) = \big(R\cos(\frac uR),R\sin(\frac uR),v)$, we find
\begin{align*}
\frac{\partial F}{\partial u} = dF_{(u,v)}\left(\frac{\partial}{\partial u}\right) &=
\big(-\sin(\frac uR),\cos(\frac uR), 0\big) = \frac1R\frac{\partial}{\partial\theta} \\
\frac{\partial F}{\partial v} = dF_{(u,v)}\left(\frac{\partial}{\partial v}\right) &= (0,0,1) = \frac{\partial}{\partial z}.
\end{align*}
Thus, the matrix representation of $dF_{(u,v)}$ in terms of the given bases is
$$\left[\begin{matrix} \frac1R & 0 \\ 0 & 1 \end{matrix}\right].$$
(It's hard to tell here, but remember that the columns of this matrix tell us how the respective basis vectors map.)
