Differential of the restriction of a function

We know that if $F:{\mathbb R}^3\rightarrow{\mathbb R}^m$ is a smooth function, and $S$ is a surface, then $F_{|S}:S\rightarrow{\mathbb R}^m$ is differentiable. Moreover, if $p\in S$, then $(dF_{|S})_p={(dF)_p}{\Big |}_{T_pM}$. In the right side, we have the differential of the restriction of $F$ and in the right side, we have the derivative of $F$ which is only viewed in the tangent plane $T_pS$. Denote $f=F_{|S}$. From now, and to simplify, we suppose $m=1$.

It is usual that we work with the Jacobian matrix of $dF$, which is noting the matrix expression of the linear map $dF$ with respect to the usual basis of Euclidean spaces, here, of ${\mathbb R}^3$ and ${\mathbb R}$. By definition, the elements are the partial derivatives of $f$, namely, $$\mbox{Jac F}(p)=M(dF_p,B_u,B_u')=(f_x\ f_y\ f_z)(x,y,z),$$ where $p=(x,y,z)$.

If we want to write the matrix expression of the linear map $df_p$, we need to fix basis in both spaces, that is, in $T_pS$ and in ${\mathbb R}$. Although for ${\mathbb R}$ we choose the basis $B_u'=\{1\}$, there is not a natural basis on $T_pS$. This only appears when we fix previously a parametrization $X=X(u,v)$ around $p$. Then the basis could be $B=\{X_u(q),X_v(q)\}$, with $X(q)=p$ and the matrix would be $M(df_p,B,B_u')$. Therefore, there is not a relation between this matrix and $\mbox{Jac F}(p)$.

We show an example. In order to avoid 'simple numbers', consider the elliptic paraboloid $z=x^2+y^2$ and the function $F(x,y,z)=z$. Then $\mbox{Jac F}(x,y,z)=(0\ 0\ 0)$. 

Consider now $f=F_{|S}$. Then $(df)_p(v)=\alpha_3'(0)=v_3$, where $v=(v_1,v_2,v_3)$. Let $p=(x,y,z)\in S$ and the parametrization $X(x,y)=(x,y,x^2+y^2)$. Then $X_x=(1,0,2x)$ and $X_y=(0,1,2y)$. Moreover, using the chain-rule, we have

$$(df)_{X(x,y)}(X_x)=(f\circ X)_x=2x,\  (df)_{X(x,y)}(X_y)=(f\circ X)_y=2y.$$

If $B=\{X_x,X_y\}$, we have  $$M((df)_{X(x,y)},B,B_u')=(2x\ 2y),$$

a matrix completely different than ${\mbox Jac F}(p)$.