I'll use a concrete example to illustrate what I want to do. Let's say I have the surfaces $Z = x^2 + y^2$ and $Z = 1$. Assume that I start at point $(1,0,1)$, which I know is in the intersection of the two surfaces. I eventually want to end up at a point that I also know is on the intersection. Again, for concreteness, let's just say I want to go to $(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},1)$.
I know that for this particular example I need to move along the unit circle $x^2 + y^2 = 1$. However, I'm just using it as an example for which I know the solution. Let's assume that I am unable to find an explicit formula for the intersection between my two surfaces. How do I choose a curve that moves from the start point to the end point that is guaranteed to always be on the intersection?
I presume we are in dimension 3. If the intersection is regular then you may use an ode to bring you around. Now, where to finish your path depends on how you specify the points. But say that the curve is the intersection of $f(z,y,z)=0$ and $g(x,y,z)=0$ and that you start at $\xi_0=(x_0,y_0,z_0)$. Then $f$ and $g$ are constant along a curve $\xi_t=(x_t,y_t,z_t)$ passing through that point, whence $\nabla f \cdot \dot{\xi}=\nabla g \cdot \dot{\xi}=0$. When gradients are independent vectors (so the intersection is regular) then $$ \dot{\xi} = C(t) \ \nabla f(\xi)\times \nabla g(\xi), \ \ \xi(0)=\xi_0$$ where the constant $C(t)$ in principle may be chosen at your convenience (also negative). Solving this ode gives you a path.