Assume that we are working on $\mathbb{R}^p$ with an inner product induced by the positive definite matrix $\mathbf{G}$, i.e. for $\mathbf{f}, \mathbf{g} \in \mathbb{R}^p$ we define $\langle \mathbf{f}, \mathbf{g} \rangle_{\mathbf{G}} = \mathbf{f}^{\top} \mathbf{G} \mathbf{g}$. This is of course a valid inner product in view of the positive-definiteness of $\mathbf{G}$. My question is, given the null-space of another matrix $\mathbf{A}$-let's call it $N(\mathbf{A})$, can we write
$$\mathbf{f} = \mathbf{f}_1 + \mathbf{f}_2,$$
with $\mathbf{f}_2 \in N(\mathbf{A}), f_1 \in N(\mathbf{A})^{\perp}$ and $\langle \mathbf{f}_1, \mathbf{f}_2 \rangle_{\mathbf{G}} = 0$? In other words, is this special but important case covered by the projection theorem? Thank you very much for your help.
Yes, there is always such a decomposition, and we don't even need to refer to a particular matrix $\bf A$. Every null space of a matrix is a subspace, and every subspace is the null space of some matrix, so we may as well just show the claim for subspaces $S$.
For any inner product $\langle \,\cdot\,,\,\cdot\,\rangle$ on $\Bbb R^p$, any subspace $S \subset \Bbb R^p$ defines an orthogonal subspace $$S^{\perp} := \{{\bf x} \in S : \langle {\bf x}, {\bf y} \rangle = 0 \textrm{ for all ${\bf y} \in S$}\} .$$ Now, nondegeneracy implies that $\dim S + \dim S^\perp = p$, and positive definiteness implies that $S \cap S^{\perp} = \{{\bf 0}\}$: For any ${\bf x} \neq {\bf 0}$ we have $\langle {\bf x}, {\bf x} \rangle > 0$, so if ${\bf x} \in S$, we must have ${\bf x} \not\in S^{\perp}$.
Thus, $\Bbb R^p$ decomposes as an orthogonal direct sum $$\boxed{\Bbb R^p = S \oplus S^{\perp}} .$$ In particular, we can decompose any element ${\bf x} \in \Bbb R^p$ uniquely as a sum $${\bf x} = {\bf x}^\top + {\bf x}^\perp, \qquad \textrm{where} \qquad {\bf x}^\top \in S, {\bf x}^\perp \in S^{\perp} ,$$ and by definition $\langle {\bf x}^\top, {\bf x}^\perp \rangle = 0$.