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$\left(\nabla^2 - \frac{1}{c^2}\frac{\partial^2}{\partial t^2}\right)\phi = \frac{m^2c^2}{\hbar^2}\phi$

with the wave function $\phi$ being a relativistic scalar: a complex number which has the same numerical value in all frames of reference. The space and time derivatives both enter to second order. This has a telling consequence for the interpretation of the equation. Because the equation is second order in the time derivative, then by the nature of solving differential equations, one must specify both the initial values of the wave function itself and of its first time derivative, in order to solve definite problems. Because both may be specified more or less arbitrarily, the wave function cannot maintain its former role of determining the probability density of finding the electron in a given state of motion. In the Schrödinger theory, the probability density is given by the positive definite expression.

Can anyone explain this quote mathematically? Why is time and space derivatives being second order so problematic?

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Specifying both $\phi({\bf x},t)$ and $\dfrac{\partial}{\partial t} \phi({\bf x},t)$ at $t=0$ determines $\dfrac{\partial}{\partial t} \int_{{\mathbb R}^n} |\phi({\bf x},t)|^2\ d{\bf x}$ at $t=0$. But if that is nonzero, $\int_{{\mathbb R}^n} |\phi({\bf x},t)|^2\ d{\bf x}$ won't be constant in time. If $|\phi({\bf x},t)|^2$ was a probability density, that integral would have to be $1$.

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The meaning of that sentence is the following. Schroedinger equation admits a conserved current such that $$ \frac{\partial}{\partial t}|\psi|^2=-\nabla\cdot\left(i\psi^*\nabla\psi-i\nabla\psi^*\psi\right). $$ So, we can integrate on a volume $V$ and obtain $\frac{\partial}{\partial t}\int_VdV|\psi|^2=0$. The quantity $\int_VdV|\psi|^2$ is positive and a probabilistic interpretation holds.

When you turn your attention to Klein-Gordon equation for a complex scalar field, the current gives the equation $$ \partial_\mu(i\phi^*\partial^\mu\phi-i\partial^\mu\phi^*\phi)=0 $$ that will yield on a finite volume $$ \frac{\partial}{\partial t}\int_VdV\left(\phi^*\frac{\partial}{\partial t}\phi-\frac{\partial}{\partial t}\phi^*\phi\right)=0 $$ but now the integral $\int_VdV\left(\phi^*\frac{\partial}{\partial t}\phi-\frac{\partial}{\partial t}\phi^*\phi\right)$ can take also negative values and a probabilistic meaning cannot be given to this. The reason is explained in quantum field theory and this integral is just the balance between positive and negative charges while you work in a second quantized framework.

I apologize with mathematicians if this seems somehow out of place but this question seems to come out from a physics textbook and, maybe, physics.stackexchange is a more correct place to post it.

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