In exterior calculus, the $k$-form Laplacian operator is
$$\Delta:=\star d \star d+ d \star d \star =\delta d+ d\delta $$
This operator applied to a $0$-form or function I believe is reduced to $\Delta:=\star d \star d.$
Say the function is $\phi,$ then $d\phi$ is the differential of the function, and the Hodge star of $d\phi$ in $\mathbb R^3$ is something that looks like
$$\text{something } dy\wedge dz + \text{something else } dz\wedge dx+ \text{other }dx\wedge dy$$
Are the coefficients just the partial derivatives of
$$d\phi=\frac{\partial\phi}{\partial x}dx+\frac{\partial\phi}{\partial y}dy + \frac{\partial\phi}{\partial z}dz$$
as in
$$\star d \phi=\frac{\partial\phi}{\partial x} dy\wedge dz + \frac{\partial\phi}{\partial y} dz\wedge dx+ \frac{\partial\phi}{\partial z}dx\wedge dy$$
?
Then applying $d$ on this last expression would produce
$$\begin{align} d\star d \phi&=\frac{\partial\phi}{\partial x}\left(\frac{\partial\phi}{\partial x}\right) dx\wedge dy\wedge dz + \frac{\partial\phi}{\partial y}\left(\frac{\partial\phi}{\partial y}\right) dy\wedge dz\wedge dx+ \frac{\partial\phi}{\partial z}\left(\frac{\partial\phi}{\partial z}\right)dz\wedge dx\wedge dy\\ & =\frac{\partial^2\phi}{\partial x^2}+ \frac{\partial^2\phi}{\partial y^2} + \frac{\partial^2\phi}{\partial z^2} dx\wedge dy\wedge dz \end{align}$$
If the preceding operations are correct, the last Hodge $\star$ would simple drop the wedges at the end of the expression to end up with a regular scalar function:
$$\star d \star d \phi=\nabla^2\phi=\frac{\partial^2\phi}{\partial x^2}+ \frac{\partial^2\phi}{\partial y^2} + \frac{\partial^2\phi}{\partial z^2} $$
If all this is correct, how would that then extrapolate to $$\Delta:=\star d \star d+ d \star d \star =\delta d+ d\delta ?$$