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.
General approach of computing a gradient
Your axis-aligned ellipsoid can be rewritten as
$$ f(x,y,z) := b^2c^2x^2 + a^2c^2y^2 + a^2b^2z^2 = a^2b^2c^2 $$
Now the gradient of that left hand side consists of the partial derivatives.
$$ \vec\nabla f = \begin{pmatrix}
2b^2c^2x \\
2a^2c^2y \\
2a^2b^2z
\end{pmatrix} $$
Sinde for the normal direction the magniture is irrelevant, you might drop that factor $2$ in each of these terms.
For a different orientation, make sure to write the ellipsoid as a polynomial in $x,y,z$. Then you can apply the same technique of computing partial differentials.
Finding the formula for rotated spheroid
If you take the spheroid $\frac{x^2+y^2}{a^2}+\frac{z^2}{c^2}=1$ and rotate it so that the original $z$ axis aligns with a vector $r=(s,t,u)$, what equation do you get? The OP actually asked this very question, and in my answer there I came up with the following equation for the spheroid:
\begin{multline*}
c^2\bigl((tz-uy)^2+(ux-sz)^2+(sy-tx)^2\bigr) + a^2(sx+ty+uz)^2 \\
= a^2c^2(s^2+t^2+u^2)
\end{multline*}
Now expand that, collect terms with common monomials, and do the partial derivatives as above. I used a bit of computer algebra here.
$$\vec\nabla f = 2\begin{pmatrix}
a^2s^2+c^2(t^2+u^2)&(a^2-c^2)st&(a^2-c^2)su\\
(a^2-c^2)st&a^2t^2+c^2(s^2+u^2)&(a^2-c^2)tu\\
(a^2-c^2)su&(a^2-c^2)tu&a^2u^2+c^2(s^2+t^2)
\end{pmatrix}\cdot\begin{pmatrix}x\\y\\z\end{pmatrix}$$
Best Answer
If $c$ is a regular value of a smooth function $f$, then $S=f^{-1}(c)$ is a surface for which $\nabla f(p)$ is normal to $T_pS$. Here $f(x,y,z) = 2x^2+y^2+2z^2$ and $p = (1,1,1)$, so the normal to the ellipsoid at $p$ is $$\nabla f(1,1,1) = (4x,2y,4z)\big|_{x=y=z=1} = (4,2,4).$$