Showing $\Delta(\Omega^p M)=d(\Omega^{p-1}M)\oplus \delta(\Omega^{p+1}M)$ Let $M$ be a compact Riemannian manifold, $*$ be the Hodge star operation, $d$ the exterior derivative, $\delta$ the codifferential $(\delta\omega=(-1)^{n(k+1)+1} *d*\omega$ where $n=\dim M$ and $k=\deg \omega$), and $\Delta=d\delta+\delta d $ the Laplacian operator. How can we show that $\Delta(\Omega^p M)=d(\Omega^{p-1}M)\oplus \delta(\Omega^{p+1}M)$? (Here $\Omega^pM$ denotes the space of $p$-forms on $M$.)
To show this, we have to show the followings:

*

*$d(\Omega^{p-1}M)\subset \Delta(\Omega^pM)$ and $\delta(\Omega^{p+1}M)\subset \Delta(\Omega^pM)$,


*$d(\Omega^{p-1}M)\cap \delta(\Omega^{p+1}M)=0$, and


*$d(\Omega^{p-1}M)+\delta(\Omega^{p+1}M)=\Delta(\Omega^pM)$.
The third one is obvious by definition of $\Delta$. The second one follows from the fact that they are orthogonal. But how can we show 1?
 A: I am safe to assume that we have a metric (which shall be denoted by $\langle\cdot,\cdot\rangle$) and orientation in the manifold so that we can talk about the Hodge star operator $\star$. Such operator is uniquely determined by the relation
$$\langle \omega,\eta\rangle\text{vol}=\omega\wedge\star\eta,$$
where $\text{vol}$ is the usual Riemannian volume form.
In that sense, we know that the adjoint of $d:\Omega^p(M)\rightarrow\Omega^{p+1}(M)$ with respect to the obvious metric induced on differential forms is given by $d^*=\delta$ (use Stokes' Theorem). Moreover, define the operator $D=d+d^*$ over $\Omega(M)$ and notice that, since it is self-adjoint,
\begin{equation}\ker D = (\text{Im}\;D)^{\perp}.\end{equation}
Now, this also shows that $\ker D=\ker \Delta$ (use $\Delta=D^2$ and the above equality) and also $\text{Im}\;D=\text{Im}\;\Delta$ (more details on that below). Putting everything together yields
$$\text{Im}\;\Delta=(\ker D)^{\perp}.$$
Now for your question, we can prove the third assertion directly: take $\omega\in\Omega^{p-1}(M)$, $\eta\in\Omega^{p+1}(M)$ and $\tau\in\ker \Delta$ and calculate
\begin{align*}\langle d\omega+d^*\eta,\tau\rangle&=\langle d\omega,\tau\rangle+\langle d^*\eta,\tau\rangle\\
&=\langle\omega,d^* \tau\rangle+\langle\eta,d\tau\rangle\\
&=\langle\omega+\eta,D\tau\rangle\\
&=0,\end{align*}
so $d\omega+d^*\eta\in(\ker\Delta)^{\perp}=\text{Im}\;\Delta$, which shows the non-trivial inclusion in 3, so the rest should be clear.

As pointed out in the comments, to prove $\text{Im}\; D=\text{Im}\;\Delta$ one must actually know beforehand that $D=d+d^*$ has closed image. This is true because $D$ is an elliptic differential operator over $\Omega(M)$. Now, every such operator is Fredholm (meaning its kernel and cokernel are finite-dimensional) and thus $D$ has closed range. If the OP is trying to understand the Hodge Decomposition Theorem, then they might be following some path where these concepts are familiar.
Now we can prove that $D$ and $\Delta$ have the same image: since $D$ has closed image, we can write
$$\Omega(M)=\ker D\oplus \text{Im}\; D.$$
The inclusion $\text{Im}\;\Delta\subset \text{Im}\;D$ is automatic ($D^2=\Delta$) whereas for the converse one, note that $\omega\in\Omega^p(M)$ is written as $\omega=\eta+D(\alpha)$, with $\eta\in\ker D$. We then have
$$
D(\omega)=D(\eta+D(\alpha))=\Delta(\alpha),$$
so $\text{Im}\;D = \text{Im}\;\Delta$ as wanted.
