Take the Neumann Laplacian on a bounded domain $-\Delta$.
We define $(-\Delta)^{\frac 12}u = \sum_{k}\lambda_k^{\frac 12}(u,w_j)_{L^2}w_j$ where $w_j$ and $\lambda_j$ are the eigenelements when $u$ is smooth.
Using this definition (and not the other ones), how do i see that $(-\Delta)^{\frac 12}$ is a nonlocal operator?
I think the answer is: we need to compute $(u,w_j)_{L^2}$ which requires knowing what $u$ is on the whole domain. BUT the ordinary Laplacian, who is NOT nonlocal, can be defined $$(-\Delta)u = \sum_{k}\lambda_k(u,w_j)_{L^2}w_j$$ and by the reasoning above this operator is also nonlocal, which is not right.
The Laplacian is local because that is how we define it. It is just a partial differential operator. It happens to be an elliptic one, so we can define its powers in a nice way as you describe, and they are all pseudodifferential operators. This means that they are pseudolocal; the singularities of $(-\Delta)^{1/2} u$ are where those of $u$ are but the smooth parts may move around.
Integer powers of the Laplacian also have a nice spectral representation and they are pseudodifferential operators, too, but they happen to be in fact differential operators. In fact, a power of the Laplacian is a differential operator if and only if it is obviously so; other powers are only pseudodifferential.
Another way to see this is that the Laplacian doesn't have any preferred direction or point. Therefore $(-\Delta)^{1/2}$ shouldn't, either. But how could this be possible for a first order differential operator, which would have to be of the form $V\cdot\nabla$ for some vector field $V$?