These methods are just applications of two different geometric ideas to help you find a normal vector to a surface. I'm sure that you or I could do some variable pushing and prove that they are compatible, but I don't know how enlightening that would be. I think the most important thing is just to understand the geometry behind each of these ideas.
When you have a parametrized surface $r(u,v) = \left< x(u,v), y(u,v), z(u,v) \right>$ and a point $(u_0,v_0)$, you can consider two cross sections of that surface. The functions $$r(u_0,v) = \left< x(u_0,v), y(u_0,v), z(u_0,v) \right>$$
$$r(u,v_0) = \left< x(u,v_0), y(u,v_0), z(u,v_0) \right>$$
define curves in three dimensions which are contained in the plane $r(u,v)$. Convince yourself that a tangent vector to any curve contained in a surface is also tangent to the surface itself. Therefore the vectors
$$ \frac{\partial}{\partial v} r(u_0,v) \big|_{v=v_0} $$
and
$$ \frac{\partial}{\partial u} r(u,v_0) \big|_{u=u_0} $$
are both tangent to the surface at $r(u_0,v_0)$. Convince yourself that if these two vectors were parallel, then $r$ wouldn't look like a curve at this point, rather than a surface, so they should not be parallel. In linear algebra terms, these vectors span the space of tangent vectors. Their cross product will yield a vector which is normal to both of them, and therefore normal to the plane. This is the definition you stated.
The other definition uses the fact that the gradient of a function at a point is perpendicular to the level surface at that point. To understand this, it is helpful to think of the lower dimensional analogy. The gradient of a function $f(x,y)$ (which defines a surface) will be perpendicular to the level curve at any point. This is geometrically obvious if $f(x,y)$ defines a plane. The level curve will be a horizontal line, and the gradient will point in the direction of greatest slope of the plane. The same logic works, in fact, for $f(x,y)$ that is not a plane because the differentiability of $f$ tells us that it behaves like a plane at any given point.
What Stokes' Theorem tells you is the relation between the line integral of the vector field over its boundary $\partial S$ to the surface integral of the curl of a vector field over a smooth oriented surface $S$:
$$\oint\limits_{\partial S} \mathbf{F} \cdot d\mathbf{r} = \iint\limits_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S} \tag{1}\label{1}$$
Since the prompt asks how to calculate the integral using Stokes' Theorem, you can find a good parametrization of the boundary $\partial S$ and calculate the "easier" integral of the LHS of (1).
Note that the boundary of $S$ is given by:
$$\partial S=\{(x,y,z)\in \mathbb R^3 : x^2 +y^2 =25, z=0\},$$
so a good parametrization to use for $\partial S$ could be:
$$\sigma :[0,2\pi] \subseteq \mathbb{R} \rightarrow \mathbb R^3$$ $$\sigma(\theta)=(5\cos(\theta),5\sin(\theta),0)$$
and finally the integral to calculate ends up being:
$$\begin{align}\oint\limits_{\partial S} \mathbf{F} \cdot d\mathbf{r}&= \iint\limits_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S} \\ &=\int_{0}^{2\pi}(0,5\cos(\theta),e^{25\cos(\theta)\sin(\theta)})\cdot(-5\sin(\theta),5\cos(\theta),0)\, d\theta,\\&=\int_{0}^{2\pi}25\cos^2(\theta)\,d\theta\end{align}$$
and from here you can use trigonometric identities to calculate the last integral.
I hope that helps!
Best Answer
The integral $\int_C F\,dr$ is a line integral and does not require the normal to a surface when evaluated directly. Use an arbitrary parametrization $[a,b]\ni t\mapsto r(t)$ of the curve $C$ and trust your calculus: $$ \int_C F\,dr=\int_a^b F(r(t))\cdot\frac{dr}{dt}\,dt\,. $$ The integrand is a dot product of two vectors. Can you proceed? Once you calculated that you can match it with the surface integral of curl $F$. This will bring you in to a position to self-answer the original question. One more thing: how many surfaces can you choose to perform that exercise? Obviously you choose one that is as simple as possible.
To approach the surface integral directly you can parametrize the surface $S$ by $$ [c,d]\times[e,f]\ni (u,v)\mapsto s(u,v)\,. $$ The normal vector to $S$ is $s_u\times s_v$ (cross product of partial derivatives). The surface element is $$ dS=\|s_u\times s_u\|\,du\,dv $$ and we use it to calculate the total area of the surface. See a concrete example here.
The unit normal is of course $$ \mathbf{n}=\frac{s_u\times s_u}{\|s_u\times s_u\|}\,. $$ Stokes says $$ \int_C F\,dr=\int_S {\rm curl\,}F\cdot\mathbf{n}\,dS\,. $$ (Take an $F$ that satisfies ${\rm curl\, F}=\mathbf{n}\,.$ Its flux is constant equal to one on $S$ and the RHS must therefore be the total area of $S\,.$) Put back the definitions to get $$\tag{1} \int_a^b F(r(t))\cdot\frac{dr}{dt}\,dt=\int_c^d\int_e^f{\rm curl}F(s(u,v))\cdot (s_u\times s_v)\,du\,dv\,. $$
The rules to remember are the following:
In formula (1) you dont't normalize the normal vector $(s_u\times s_v)$ because $\|s_u\times s_v\|$ has cancelled out and $s_u\times s_v$ needs to scale with the parametrization to ensure independence of that integral under a re-parametrization (change of variables).
When your problem has enough symmetry to see that curl$\,F\cdot\mathbf{n}$ is constant on $S$ then you just have to figure out the surface of $S$ and multiply it with that constant. In that case you normalize $\mathbf{n}$ of course.
In your example $$ F=\begin{pmatrix} 2y\\-x\\y-x\end{pmatrix}\,,\quad S=\{x+y+z=6\} $$ the second method is smarter:
To do all calculations required in (1) explicitly I suggest a simpler example wich contains all essentials: $$ F=\begin{pmatrix}y\\-x\\0\end{pmatrix}\,,\quad S=\{z=0\}\,,\quad C=\{x^2+y^2=4\}\,. $$