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I have the equation \begin{align} \begin{cases} u_t(\boldsymbol{x},t)-\nabla^2u(\boldsymbol{x},t) = 0, &\boldsymbol{x}\in D, & t>0, \\ u(\boldsymbol{x},t)=0, & \boldsymbol{x}\in\partial D, & t>0, \\ u(\boldsymbol{x},0)=u_0(\boldsymbol{x}), &\boldsymbol{x}\in D, \end{cases} \end{align} where $\boldsymbol{x}=(x_1,x_2,x_3)$.

Does the Laplace operator acts on the vector $\boldsymbol{x}$ AND $t$? I.e. \begin{align} \nabla^2 u(\boldsymbol{x},t)=\frac{\partial^2 u}{\partial x_1^2}+\frac{\partial^2 u}{\partial x_2^2}+ \frac{\partial^2 u}{\partial x_3^2}+\frac{\partial^2 u}{\partial t^2} \end{align} or just \begin{align} \nabla^2 u(\boldsymbol{x},t)=\frac{\partial^2 u}{\partial x_1^2}+\frac{\partial^2 u}{\partial x_2^2}+ \frac{\partial^2 u}{\partial x_3^2} \end{align} If the second one is correct, why is it so?

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    Usually it is just in the spatial variables, but it depends on the context. Here you have the heat equation, so the Laplacian should be interpreted in the second form (only in space variables). If your question is why the heat equation has this form, then you should read about the derivation of the heat equation form conservation of heat and Fourier's law. – 2017-02-24
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    You should read about the D'Alembertian operator https://en.wikipedia.org/wiki/D'Alembert_operator – 2017-03-03

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In this context (heat equation) the Laplacian is taken in the space variables only -- because this is what the heat equation is.

When considering a PDE where the role of Laplacian is not clear, the author will either state the interpretation of Laplacian in text, or use indices to clarify it: $$\nabla_x^2 u(x,t) \quad \text{vs.}\quad \nabla_{x,t}^2 u(x,t)$$ or $$\Delta_x u(x,t) \quad \text{vs.}\quad \Delta_{x,t} u(x,t)$$