If virtual photons are just mathematical constructs, what is the real physical principle that allows EM field to give momentum to a charged particle? If virtual photons are just mathematical constructs, what is the "real" (if such an adjective makes sense) physical principle that allows electromagnetic field to deliver momentum to a charged particle?
 A: tl;dr See the bold paragraph below.
Your question is how the vector field $A^\mu$ "really" causes an electromagnetic force. But let's start with a toy problem: how would a scalar field $\varphi$ cause a force?
I'll work in $+---$ with $c=\hbar=1$, and denote $\varphi$'s source current by a scalar $J$. The basic idea is a brief timeslice has probability amplitude of the form$$e^{iW}=\int\mathcal{D}\varphi\exp(iS)=\langle0|e^{-iHT}|0\rangle=e^{-iET},$$where$$S=\int d^4x\left[J(x)\varphi(x)-\frac12\varphi(\square+m^2)\varphi\right]$$is an action over the timeslice, not all of spacetime. After some Fourier transforms (see e.g. Quantum Field Theory in a Nutshell Secs. I.2-I.4 for the calculation), one unit "charge" at $\vec{x}_1$ gives another at $\vec{x}_2$ the potential energy$$E=-\int\frac{d^3\vec{k}}{(2\pi)^3}\frac{\exp i\vec{k}\cdot(\vec{x}_1-\vec{x}_2)}{\vec{k}^2+m^2}=-\frac{e^{-mr}}{4\pi r}<0,$$implying an attractive force. But we need to talk about how $W$ looks, for an arbitrary matter distribution:$$W=-\frac12\int\frac{d^4k}{(2\pi)^4}\frac{|J(k)|^2}{k^2-m^2+i\varepsilon},\,J(k):=\int d^4xJ(x)e^{-ik\cdot x}.$$The $\varepsilon\to0$ later; it's just a way of being careful with the complex analysis. But notice the integral is over all of $k$-space, so it looks like $W$ is caused by particles of every $4$-momentum, including the unphysical ones with $k^2-m^2\ne0$. That's where the idea of "virtual particles" comes from, but they're symptoms of the field $\varphi$, not observable particles, which are often described as "bumps on the field".
Fields, not particles, are fundamental; "virtual particles" are just a description of the mathematical appearance of an odd way fields work in $W$. But the real physical principle by which $\varphi$ causes a potential energy $E$ is that it causes a $k$-space integral that gives the phase $W$ of the path integral.
What about the vector case, then? It's very similar (ibid. I.5, which gets the massive case, then takes $m\to0$ for electromagnetism's photon when we're ready):$$W=-\frac12\int\frac{d^4k}{(2\pi)^4}\frac{\overline{J^\mu(k)}(m^{-2}k_\mu k_\nu-g_{\mu\nu})J^\nu(k)}{k^2-m^2+i\varepsilon}.$$The charge conservation law $k_\nu J^\nu(k)=0$ simplifies this to$$W=\frac12\int\frac{d^4k}{(2\pi)^4}\frac{\overline{J^\mu(k)}J_\mu(k)}{k^2-m^2+i\varepsilon}.$$The $-$ sign on $g_{\mu\nu}$ has ensured like charges repel this time. The general rule is a spin-$s$ field attracts (repels) like charges for $s$ even (odd).
A: If you are asking at the elementary particle level, light is  a sum of a large number of real photons, by real meaning the photons have a four vector which has an invariant mass of zero, i.e. on mass shell. These photons carry momentum and can transfer a dp/dt in interacting.
The interactions of a photon with a particle or field are mathematically modeled with Feynman diagrams( second page in link). That is where virtual particles are presumed to exit, they are the carriers of the quantum numbers that are needed for correct conservation, but the mass of  their  four vectors is not zero, as it should be for the photon. The virtual particle, photon in your question,  is off mass shell. It is named as a "photon" for convenience in keeping track of the interactions.
My answer here may help in understanding photons and fields.
A: Coulomb's Law:
$$ {\bf F} = \frac 1 {4\pi\epsilon_0} \frac{q_1q_2}{r^2}{\bf \hat r}$$
There's also some magnetism, depending on spin an kinematics.
A: Heuristic explanation: We know that there is no instantaneous action at distance, two electrons for example in interaction must undergo a variation of the parameters of their environments caused by the presence of the other, if they are separated by a distance d, the variation of the parameters happens in a time $t+d/c$ (a time of delay to act), we know that for a central field (Newton's law) $\mathbf{F}dt=-\mathbf{\nabla}U dt=dp$, that we can transform in $\Delta U \Delta t= \Delta x \Delta p $, but according to Heisenberg's uncertainty principle : $\Delta U \Delta t= \Delta x \Delta p\geq \hbar $.
A violation of the law of conservation of energy up to a quantity $\Delta E=\Delta U$(change in the energy of the environment brought by the energy of the photo) will be hidden by the uncertainty on the energy provided that the time available to make the observation $\Delta t$ is sufficiently large such that $\Delta t = \frac{\hbar}{\Delta U }$, if the emissaries of the one towards the other to prevent him of its presence in its environment are photons of energy $h\nu$ and that $\Delta t =\frac{\hbar}{ h\nu }$ .
These photons are then observable on a maximum distance of: $R= c \Delta t=\frac{c}{4\pi\nu }<d$,
and as the frequency can be arbitrarily small, the range of the force transmitted by the massless photon is unlimited. It may appear from this relation that the range is limited for a free photon. But this would be forgetting that a free photon does not exist, because it would have a totally undetermined frequency. So the interaction distance would be too.
These exchange quanta, which are unobservable, are called "virtual photons". As photons are not charged, we also say that the interaction is carried out by "neutral current".
A much more satisfactory approach is to use mass as the energy term:  $\Delta t \le \frac{\hbar}{mc^{2} }$
Using this relation, it is possible to know the time during which a virtual particle can travel a distance that would correspond to:$ R= c \Delta t= \frac{\hbar}{mc^{2} }$ which intervenes in the  Yukawa potential.
