For the first part of your question: indeed, if you define $L$ like that, the expression $L \alpha = \frac{\partial^2}{\partial x^2} \alpha + \frac{\partial^2}{\partial y^2} \alpha$ is correct. What is not so right is saying that "$L$ maps $\alpha$ to itself - it doesn't, in the same sense the derivative of a function is a new function, usually different from the original function.
Which brings me to the next point: I'm not too sure of how you interpret the term "variable" here, but in this case $\alpha$ must be a function, and not a regular variable. The idea behind working with operators like this is not so much talking about the operation being performed, but rather talking about properties the operator as such might have. It is, in a way, a generalization of finite-dimensional linear algebra. There you have a linear equation, say $Tx = y$, and you are concerned with what the eigenvalues of $T$ look like and how they determine the kind of possible solutions to that equation. It's similar here, you can still find the eigenvalues of the Laplacian. But the function spaces we deal with here are infinite-dimensional.
For the second part of your question: what is $p$? Is $\alpha$ still understood to be a function, as in the previous part?