Why the second-degree Laplace operator is the sum of the $x$ and $y$ components
$$\nabla^2 f = \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2}$$
but the first-degree Laplace operator is a vector of $x$ and $y$ components?
$$\nabla f = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right)$$
instead of just
$$\nabla f = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y}$$
One way to derive the Laplacian (Laplace operator) is to view $\nabla^2$ as $\nabla\cdot \nabla$, where $\nabla=\left(\frac{\partial}{\partial x},\frac{\partial}{\partial y}\right).$
With this view , we have
\begin{align} \nabla\cdot \nabla&=\left(\frac{\partial}{\partial x},\frac{\partial}{\partial y}\right)\cdot\left(\frac{\partial}{\partial x},\frac{\partial}{\partial y}\right)\\ &=\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2} \end{align}
as inner product. Note that this means, in some sense, taking Laplacian on a function $f$ is the same as taking gradient first and then taking divergence.