The correct setting for this construction turns out to be *projective* varieties, so let me suppose we have a smooth variety $X$ inside $\mathbf{P}^N$, for some $N \ge 1$, defined by the vanishing of some homogenous polynomials $F_1, \dots, F_r$ in variables $x_0, \dots, x_N$, with the $F_i$ having coefficients in $\mathbf{Q}$. Actually, let me assume the $F_i$ have coefficients in $\mathbf{Z}$, which is no loss since we can just multiply up. Then we can make sense of the reduction of $X$ modulo $p$; and we want to study the point counts $\#X(\mathbf{F}_p)$ as a function of $p$, possibly neglecting some finite set $\Sigma$ containing all primes such that the reduction of $X$ mod $p$ is singular.

Thanks to Grothendieck, Deligne, and others, we have a very powerful bunch of tools for analysing this problem. The setup is as follows. Choose your favourite prime $\ell$. Then the theory of **etale cohomology** attaches to $X$ a bunch of finite-dimensional $\mathbf{Q}_\ell$-vector spaces $H^i_{\mathrm{et}}(X_{\overline{\mathbf{Q}}}, \mathbf{Q}_\ell)$ (let me abbreviate this by $H^i_\ell(X)$ to save typing). The dimension of $H^i_\ell$ is the same as the $i$-th Betti number of the manifold $X(\mathbb{C})$; but they encode much more data, because each $H^i_\ell(X)$ is a representation of the Galois group $\operatorname{Gal}(\overline{\mathbf{Q}} / \mathbf{Q})$, unramified outside $\Sigma \cup \{\ell\}$; so for every prime not in this set, and every $i$, we have a number $t_i(p)$, the trace of Frobenius at $p$ on $H^i_\ell$, which turns out to be independent of $\ell$.

**Theorem**: $\#X(\mathbf{F}_p)$ is the alternating sum $\sum_{i=0}^{2 dim(X)} (-1)^i t_i(p)$.

Now let's analyse $H^i_\ell$ as a Galois representation. Representations of Galois groups needn't be direct sums of irreducibles, but we can replace $H^i_\ell$ by its semisimplification, which does have this property and has the same trace as the original $H^i_\ell$. This semisimplification will look like $V_{i, 1} + \dots + V_{i, r_i}$ where $V_{i, j}$ are irreducible; and the $V_{i, j}$ all have motivic weight $i$, so the same $V$ can't appear for two different $i$'s. So we get a slightly finer decomposition

$\#X(\mathbf{F}_p) = \sum_{i=0}^{2 \mathrm{dim} X} (-1)^i \sum_{j=1}^{k_i} t_{i, j}(p)$

where $t_{i,j}(p)$ is the trace of $Frob_p$ on $V_{i,j}$.

Let me distinguish now between several different types of irreducible pieces:

- $V_{i, j}$ is a one-dimensional representation. Then $i$ must be even, and the trace of Frobenius on $V_{i, j}$ is $p^{i/2} \chi(p)$ where $\chi$ is a Dirichlet character.
- $V_{i, j}$ is two-dimensional and comes from a modular form. Then $t_{i,j}(p) = a_p(f)$, and $f$ must have weight $i+1$.
- $V_{i,j}$ is two-dimensional and doesn't come from a modular form. This can happen, but it's rare, and it's expected that all examples come from another kind of analytic object called a Maass wave form; in particular this forces $i$ to be even.
- $V_{i, j}$ has dimension $> 2$. Then $V_{i, j}$ cannot be the Galois representation coming from a modular form, because these always have dimension 2.

You seem to want your varieties to have $X(\mathbf{F}_p)$ = (polynomial in $p$) + (coefficient of a modular form). From the above formulae, it's clear that this can only happen if all the $V_{i,j}$ have dimension 1 or 2; there is exactly one with dimension 2 and it comes from a modular form; and the one-dimensional pieces all come from the trivial Dirichlet character. This always happens for genus 1 curves, because the $H^0$ and $H^2$ are always 1-dimensional for a curve, and the genus condition forces the $H^1$ to be two-dimensional.

However, once you step away from genus 1 curves, this is totally not the generic behaviour, and it will only occur for unusual and highly symmetric examples, such as the rigid Calabi-Yaus and extremal $K_3$ surfaces in the links you've posted.