Question: Let M a riemannian manifold. How can I show that the hodge-laplace-operator of a function $f$ is the negative of the laplace-operator?

Answer: 
A rather short proof can be found [here](http://en.wikipedia.org/wiki/Proofs_involving_the_Laplace%E2%80%93Beltrami_operator).

I assume you are interested in the case when $f$ is a *scalar* function. Otherwise the
Hodge Laplacian differs from the Laplace–Beltrami operator not only by a sign due to the Ricci curvature. See the [Weitzenböck identity](http://en.wikipedia.org/wiki/Weitzenb%C3%B6ck_identity).
For the function case, choose local coordinate $(x\_1,\cdots, x\_n)$, then $\nabla\_i\nabla\_j f=Hessf(\frac{\partial}{\partial x\_i}, \frac{\partial}{\partial x\_j})=\frac{\partial^2 f}{\partial x\_i\partial x\_j}-\Gamma^k\_{ij}f\_k$, and
\begin{equation} 
\Delta f =g^{ij}\nabla\_i\nabla\_j f=\frac{1}{\sqrt{g}}\frac{\partial}{\partial x\_i}(\sqrt{g}g^{ij}\frac{\partial f}{\partial x\_j}),
\end{equation}
where $g=\det{g\_{ij}}$.

Next, calculating $\Box f=d^\*d f$. First, for $\alpha\in A^1(M)$, we calculate $d^\*\alpha$.
Let $\alpha=\alpha\_i dx^i$, $\beta\in C^{\infty}(M)$, let $d^\*\alpha=A$, then,
\begin{equation}
\begin{split}
\int\_X<d^\*\alpha, \beta>dV&=\int\_X<\alpha, d\beta>dV\\
&=\int\_X \alpha\_i\partial\_j\beta g^{ij}dV\\
&=-\int\_X\frac{\partial}{\partial x\_j}(\alpha\_i g^{ij}\sqrt{g}) \sqrt g\beta dV
\end{split}
\end{equation}
So, $d^\*\alpha=-g^{-\frac{1}{2}}\frac{\partial}{\partial x\_j}(\alpha\_i g^{ij} g^{\frac{1}{2}})$, and 
\begin{equation}
\begin{split}
\Box f &=d^\*d f=d^\*(f\_i dx^i)\\
&=-\frac{1}{\sqrt{g}}\frac{\partial}{\partial x\_i}(\sqrt{g}g^{ij}\frac{\partial f}{\partial x\_j})=-\Delta f.
\end{split}
\end{equation}