How to reconcile two different intuitions of the Laplacian operator? In here the Laplacian $u_{xx}+u_{yy}$ is said to:

The Laplacian measures the degree by which the value at a point differs from the average of its neighbors.

whereas here the intuition departs from a more stringent set up: there is a scalar-valued function $f(x,y),$ from which the Laplacian is defined as the divergence of the gradient, $\nabla\cdot\nabla f.$ The intuition is:

A measure of how much of a minimum a point is in a scalar valued multivariate function.

The question is where the average appears in the first interpretation - is it truly the average as understood in statistics? Because the second derivative in different directions seems to directly speak about the curvature of the graph - there is no averaging.
And is the gradient also implied in the first interpretation? Or there is no real need for a gradient to understand the Laplacian?
 A: Consider the square centered at $(0,0)$ with coordinates $(\pm h,\pm h)$.
Average the value of $u$ at these four corners and measure the difference from $u(0,0)$:
$$
    \frac{u(h,h) + u(-h,h) + u(h,-h) + u(-h,-h)}{4} - u(0,0)
    = \frac{u(h,h) + u(-h,h) + u(h,-h) + u(-h,-h) - 4u(0,0)}{4}
$$
by Taylor's theorem, 
\begin{align*}
    u(h,h) &= u(0,0) + h u_x(0,0) + h u_y(0,0) + \frac{1}{2}h^2(u_{xx}(0,0) + 2u_{xy}(0,0) + u_{yy}(0,0)) +R(h,h) \\\\
    u(h,-h) &= u(0,0) + h u_x(0,0) - h u_y(0,0) + \frac{1}{2}h^2(u_{xx}(0,0) - 2u_{xy}(0,0) + u_{yy}(0,0)) +R(h,-h) \\\\ 
    u(-h,h) &= u(0,0) - h u_x(0,0) + h u_y(0,0) + \frac{1}{2}h^2(u_{xx}(0,0) - 2u_{xy}(0,0) + u_{yy}(0,0)) +R(-h,h) \\\\
    u(-h,-h) &= u(0,0) + h u_x(0,0) - h u_y(0,0) + \frac{1}{2}h^2(u_{xx}(0,0) + 2u_{xy}(0,0) + u_{yy}(0,0)) +R(-h,-h) \\\\ 
\end{align*}
You see that if we add these four together,
$$
\frac{u(h,h) + u(-h,h) + u(h,-h) + u(-h,-h) - 4u(0,0)}{4}
= h^2 \left(u_{xx}(0,0)+u_yy(0,0)\right) + \text{remainders}
$$
So
$$
   \lim_{h\to 0} \frac{u(h,h) + u(-h,h) + u(h,-h) + u(-h,-h) - 4u(0,0)}{4h^2} = \nabla^2 u(0,0)
$$
