Let me consider the surface $S:=\{(x,y,z): x^2+y^2=1\}$ and $G:=\{(0,y,z): y,z\in \Bbb{R}\}$ and define $f:C\rightarrow G$ by $f(x,y,z)=(0,y,z)$. Let us take the following two patches for $C$ and $G$: $$\sigma_1(u,v)=(\cos(u),\sin(u), v)$$and respectively $$\sigma_2(w,z)=(0,w,z)$$ Now I want to compute the differential at an arbitrary point $p\in C$ and express it in the base of the tangent plane.
My idea was the following:
Let me pick an arbitrary point $p\in C$, then since $\sigma_1$ is a patch we know that $p=\sigma_1(u_0,v_0)$ for some $u_0,v_0$. Then let us define $$\begin{align}\gamma(t)&=\sigma_1(u_0+t,v_0)\\\beta(t)&=\sigma_1(u_0,v_0+t)\end{align}$$ we see that $\gamma\perp \beta$. At this point also remark that $\gamma'(0)=\sigma_{1,u}(u_0,v_0)$ and $\beta'(0)=\sigma_{1,v}(u_0,v_0)$
Then let us compute $$\begin{align}Df_p(\gamma'(0))&=(f\circ\gamma)'(0)\\&=\frac{d}{dt}~\left(f(\gamma(t))\right)\big|_{t=0}\\&=(0,\cos(u_0),0) \end{align}$$ and similarly $$\begin{align}Df_p(\beta'(0))&=(f\circ\beta)'(0)\\&=\frac{d}{dt}~\left(f(\beta(t))\right)\big|_{t=0}\\&=(0,0,1) \end{align}$$ so we have found two tangent vectors which are linealy independent. Now since the differential is linear we know that for a general curve $\alpha:[-1,1]\rightarrow C$ with $\alpha(0)=p$ and $\alpha'(0)=a\sigma_{1,u}(u_0,v_0)+b\sigma_{1,v} (u_0,v_0)$ we have $$\begin{align}Df_p(\alpha'(0))&=Df_p(a\cdot\sigma_{1,u}(u_0,v_0)+b\cdot\sigma_{1,v} (u_0,v_0))\\&=a\cdot Df_p(\sigma_{1,u}(u_0,v_0))+b\cdot Df_p(\sigma_{1,v}(u_0,v_0))\\&=a\cdot Df_p(\gamma'(0))+b\cdot Df_p(\beta'(0))\\&=a\cdot (0,\cos(u_0),0)^T+b\cdot (0,0,1)^T\\&=a\cdot \cos(u_0) \sigma_{2,w}+b\cdot \sigma_{2,v}\end{align}$$
Does this work or am I wrong?
You've done some of the correct algebra. Let's go through this, wrapping some words in various places to nail down what's going on.
First of all, can I suggest that we use different symbols for the local coordinates on $C$ and $G$? Let's use this notation: $$ \sigma(u, v) = (\cos(u), \sin(v), v) \in C$$ $$ \rho(w, z) = (0, w, z)$$
In terms of these local coordinates, the map $F$ acts like so: $$ F : (u, v) \mapsto (w, z) = (\sin (u), v).$$
(I'm abusing notation slightly, using $F$ to denote the mapping in the local coordinates.)
The differential $dF_p$ at the point $p = (u_0,v_0)$ is a linear map that takes tangent vectors in the tangent space at $p$, and returns tangent vectors in the tangent space at $f(p)$.
A tangent vector in the tangent space at $p$ takes the form $a \partial_u + b \partial_v$, where $a$ and $b$ are some real coefficients. $dF_p$ sends this vector to a vector in the tangent space at $f(p)$, which has the form $c\partial_w + d\partial_z$, where $c$ and $d$ are given by $$ \begin{bmatrix} c \\ d \end{bmatrix} = \begin{bmatrix} \frac{\partial w}{\partial u}(u_0,v_0) & \frac{\partial w}{\partial v}(u,v) \\ \frac{\partial z}{\partial u}(u_0,v_0) & \frac{\partial z}{\partial v}(u,v)\end{bmatrix} \begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} \cos(u_0) & 0 \\ 0 & 1\end{bmatrix} \begin{bmatrix} a \\ b \end{bmatrix} .$$
Edit: Perhaps you're wondering where that matrix $\begin{bmatrix} \frac{\partial w}{\partial u}(u,v) & \frac{\partial w}{\partial v}(u,v) \\ \frac{\partial z}{\partial u}(u,v) & \frac{\partial z}{\partial v}(u,v)\end{bmatrix} $ came from. If so, I'll derive the expression for the differential in another way...
The logic is basically what you've outlined. Consider the curve $$ \alpha : t \mapsto (u, v) = (u_0 + at, v_0 + bt).$$ At time $t = 0$, this curve is at position $(u, v) = (u_0, v_0)$, and its velocity vector is $a\partial_u + b\partial_v$.
Now consider the image of our curve under the map $F$. This is the curve $$ F \circ \alpha : t \mapsto (w, v) = (\sin(u_0 + at), v_0 + bt).$$ At time $t = 0$, this curve is at position $(w, v) = f(u_0, v_0) = (\sin(u_0), v_0)$.
Its velocity vector at time $t = 0$ is $$ \left( \frac{d}{dt}\sin(u_0 + at)|_{t=0}\right) \partial_w + \left(\frac{d}{dt}(v_0 + bt)|_{t=0}\right)\partial_z = a\cos(u_0) \partial_w + b\partial_z$$
Thus the differential $dF_p$ is the linear map $$ dF_p(a \partial_u + b \partial_v) = a\cos(u_0)\partial_w + b\partial_z.$$