Ok, so the intuitive pictorial idea of a derivative of a function $f$ at a point $x$ is taking the slope of the graph of $f$ at the point $x$. Now this is fine for guiding intuition but it isn't mathematically rigorous so let's try to be more rigorous.
So firstly what we do is we try to approximate the slope at a point, you may have heard of secant lines? That is what we are doing here. So to approximate the slope we take a small value $h$ and we take the slope between $f(x)$ and $f(x+h)$. Remember that gradient is "rise over run?" so our approximation is our rise $f(x+h)-f(x)$ over our run $(x+h)-x=h$. Hence our approximation is:
$$
\frac{f(x+h)-f(x)}{h}
$$
I hope you recognize this equation.
Ok so we have an approximation, now the idea is that if our function $f$ is smooth enough at $x$ then as $h$ gets smaller, this approximation is going to get better. If this is the case wouldn't we want to take $h=0$ to get the best approximation? Unfortunately we have a problem here, if you substitute $h=0$ into the approximation above we get $\frac{0}{0}$ which is bad. So what are we going to do?
Well what we do is hope that as $h$ gets close to 0, that our approximation starts to settle down and get close to the gradient we are trying to approximate. To this end we ask does
$$
\lim_{h\rightarrow 0}
\frac{f(x+h)-f(x)}{h}
$$
Exist? If it does, then our approximation is good, and we say that $f$ is differentiable at $x$. And we call the value $
f'(x)=\lim_{h\rightarrow 0}
\frac{f(x+h)-f(x)}{h}
$ the derivative of $f$ at $x$.
If the limit does not exist then it means that our function is not smooth enough to take good approximations of the gradient, and if we can't approximate it well we say it doesn't exist, and we say $f$ is not differentiable at $x$.
So a function is differentiable if the limit above exists, and the derivative is the value of the limit.
This clear things up?
First of all, when dealing with more than two variables level set is a better denomination than level curve (or level surface in three dimensions.)
Now to your question. Let $x_0\in L(c)$ and let $\gamma\colon(-a,a)\to \mathbb{R}^n$ be a $C^1$ curve contained in $L(c)$ and such that $\gamma(0)=x_0$. Then
$$
f(\gamma(t))=c,\quad -a<t<a.
$$
Differentiating with respect to $t$ and evaluating at $t=0$ we get
$$
\nabla f(x_0)\cdot\gamma'(0)=0.
$$
The set of all vectors $\gamma'(0)$ for all possible curves $\gamma$ forms the tangent hyperplane to $L(c)$ at $x_0$, and $\nabla f(x_0)$ is orthogonal to all of them, that is, the gradient is orthogonal to the tangent hyperplane of the level set.
Best Answer
Say you are standing on the side of a hill. Imagine somewhere beneath the hill, there is a flat $x,y$ plane that you can use to determine your position. Let's say $+x$ is east and $+y$ is north.
If the hill is smooth, then the height of the hill above this plane is some continuous function $f(x,y)$.
The gradient of $f$ at any point tells you which direction is the steepest from that point and how steep it is. To find the direction of the gradient of $f$ where you are standing, decide which direction is the steepest. The answer could be "north" or "30 degrees west of south". There is no vertical component to the gradient, it is telling you a direction with respect to the $x,y$ plane which is your reference. The magnitude of the gradient will be the slope of the hill in that direction.
The tangent plane is the plane that best approximates the shape of the hill where you are standing. The hill may be curved if you look at it from a distance, but maybe directly beneath your feet it is flat enough to set a pizza box down and have it be flush with the ground. The plane that the bottom of the pizza box defines would, roughly, be the "tangent" plane.