QFT and its notations

The quick answer is that (3) written by physicists is not to be taken too seriously by mathematicians. However, it is a statement of a goal or research problem which is to find a rigorous definition/construction of what (3) is trying to say.

One has an injective continuous linear map $\iota:\mathscr{S}(\mathbb{R}^d)\rightarrow \mathscr{S}'(\mathbb{R}^d)$ given by $$ \iota(\varphi)=\left(f\longmapsto \int_{\mathbb{R}^d}\varphi(x)f(x)\ d^dx\right) $$ where $f$ is a generic element of $\mathscr{S}(\mathbb{R}^d)$. Via this map it is natural to identify $\mathscr{S}(\mathbb{R}^d)$ with a subset of $\mathscr{S}'(\mathbb{R}^d)$. Moreover, this subset is dense (in fact sequentially dense) in $\mathscr{S}'(\mathbb{R}^d)$. The action $S(\varphi)$ in (2) is perfectly well defined for $\varphi$ in the subset $\mathscr{S}(\mathbb{R}^d)$ but not for $\varphi\in \mathscr{S}'(\mathbb{R}^d)\backslash\mathscr{S}(\mathbb{R}^d)$.

Unfortunately, the Gaussian measure $d\mu_G$ is not supported on $\mathscr{S}(\mathbb{R}^d)$ but on the much bigger space $\mathscr{S}'(\mathbb{R}^d)$. One can be a bit more precise and work in a weighted Sobolev or Besov space of exponent $\alpha$ but the latter would be negative except for $d=1$.

So one has to introduce a regularization in order to remove the regularization. This is explained in my answer to

A roadmap to Hairer's theory for taming infinities

and to

https://physics.stackexchange.com/questions/372306/what-is-the-wilsonian-definition-of-renormalizability

To make an interacting (i.e. not purely quadratic) QFT at all meaningful, you have to impose a regulator. The most transparent regulator in many ways, and the only known regulator that allows to address the full non-perturbative content of the QFT, is a lattice regulator. Replacing $\mathbb{R}^d$ by $\mathbb{Z}^d$ (or a finite subset such as $\Lambda=\mathbb{Z}^d\cap[0;L]^d$ with suitable boundary conditions) makes the path integral measure $\mathrm{d}\varphi=\prod_{x\in\Lambda}\mathrm{d}\varphi_x$ unambiguously defined and removes any problems with pointwise operations on the fields appearing inside $S(\varphi)$. Whether a continuum limit then exists is of course a non-trivial (and generally unsolved) problem.

To answer your explicit question: Quantum field theorists (if such a generalization is permitted), at least when writing expressions such as the action of $\varphi^4$ theory, tend to think not so much in terms of distributions as in terms of fields in the regularized theory, for which pointwise products are unproblematic.

It should also be noted that a QFT is the quantization of a classical field theory, and that is governed by an action that involves pointwise products; I'd consider that sufficient motivation to keep the pointwise notation (which moreover makes the Poincaré invariance of the continuum action apparent, something which a more distribution-centric approach would likely obscure).