The comments give one way of understanding, but I figured I could maybe explain the original proof better.
Suppose our plane has normal vector $n$ (which is normal to the plane at every point on it) and an example point on the plane: $p$. These two vectors, one a direction and the other a point, determine any plane.
Then, our plane equation is $$ n\cdot(x-p)=0 $$
that is, every $x$ that satisfies this equation is a member of the plane. This is equivalent to $n\cdot x = n\cdot p$.
Notice that $x=p$ is a member of the plane. But, $x=p+n$ is not a member, nor is $x=\vec{0}$, assuming $p\ne \vec{0}$.
Another way to write this is as a point set $\Pi$:
$$
\Pi = \{ \;x\in\mathbb{R}^3\;|\;n\cdot x= n\cdot p\;\}
$$
i.e. these are the set of points making up the plane. Again, notice that if $p=\vec{0}$, then the origin is a point in $\Pi$. But if $p$ does not vanish, then the origin is not in the plane.
The reason is that $p$ is a translation or shifting parameter. That is, a plane has an orientation parameter, $n$, which "rotates" it, and a position parameter $p$, which slides the plane around. When $p=\vec{0}$, we have slid the plane so that it intersects that origin. The plane equation in this case is $n\cdot x = 0$.
Here's a different approach. Every plane is determined by giving 3 unique points. Let's take $p,a,b$. Define $T_1=a-p$ and $T_2=b-p$. We can suppose $T_1$ and $T_2$ are orthogonal; if they are not, we can use Gram-Schmidt orthonormalization.
Now, suppose we walk around between $p$ to $a$ or $b$. This is the same as adding some multiple of $T_1$ or $T_2$ to $p$. So a parametric equation for the plane is
$$
x(s,t) = p + sT_1 + tT_2
$$
so that if you input any $s$ and $t$, your output is a point on the plane.
See also here.
Notice that a normal vector is simply $n=(T_1\times T_2)/||T_1\times T_2||_2$.
So we get equivalence to $n\cdot(x-p)=0$ as before.
A little bit more explicit:
Normal to the plane: $n = (1,3,1)$.
Line through $a = (1,2,6)$ :
$r(t) = (1,2,6) + t (1,3,1)$.
$x(t) = 1 + t; $ $y((t) = 2 + 3t;$
$z(t) = 6 + t$;
Plane and line intersect:
$x(t) + 3y(t) + z(t) = 4$, an equation for $t$.
Solve for $t$ and use this $t = t_0$ value in
$r = (1,2,3) + t_0 ( 1,3,1)$
to find the coordinates of the point of intersection.
Best Answer
Yes, you are correct. Only 'but' is that the vector should be on the $n_i$'s, not on the $1$ (and you don't really need the $1$ there), so the last equation should read: $$f'(x)=\sum_{i}^{} (n_i) = \vec{n}.$$
Think that the function $f$ has planes parallel to this one as its level sets. This means that $f$ doesn't change on the planes, and so the direction in which is grows the most is the one perpendicular to the planes. I hope you find this intuition helpful in the future.
Addendum: remaking calculations with more rigorous notation.
We have the plane defined as the zero set of a function $f:\mathbb R^3\to \mathbb R$. This is, $$f(x,y,z) := (\vec x - \vec p)\cdot\vec n = [(x,y,z) - (p_x,p_y,p_z)]\cdot (n_x,n_y,n_z).$$
The gradient of $f$ is the vector whose $n$-th coordinate is the derivative of $f$ in the direction of that coordinate. It is a vector field $\nabla f:\mathbb R^3\to \mathbb R^3$ associating to each point in the space a vector that is tangent to it. In this case, where $$f(x,y,z) = (x-p_x)n_x + (y-p_y)n_y + (z-p_z)n_z, $$ the gradient shall be $$\nabla f (x,y,z) = \left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z}\right) = (n_x,n_y,n_z) = \vec n. $$