Let me try this more clearly than the other answer, which aren't wrong. You ask:

So, can someone please elaborate what this EM field is with respect to $\vec E$ and $\vec B$ in the context of Helmholtz decomposition?

There is no "EM field in the context of Helmholtz decomposition". Helmholtz just says that every vector field $\vec V$ is decomposable as curl and gradient of two other fields, i.e.

$$\vec V = \vec \nabla \phi + \vec \nabla \times \vec A $$

You can do this for the electric or the magnetic field, of course, but this isn't particularly enlightening as to the nature of "the EM field". A field should behave nicely under transformations, and special relativity with its action on the electric and magnetic fields shows us that we should not add them together, but seek a quantity that transforms nicely under Lorentz transformations instead:

"The electromagnetic field" is equivalently the gauge four-potential $A$ (consisting of the scalar electrostatic potential in the temporal and the magnetic vector potential in the spatial entries) or its derivative, the field strength tensor $F = \mathrm{d}A$. Electric and magnetic fields become part of the tensor as \begin{align} F^{0i} & = E^i \\ F^{ij} & = \sum_k\epsilon^{ijk}B^k \end{align} This is "the EM field", but it has nothing to do with Helmholtz decomposition, since electromagnetism is properly looked at in the four-dimensional setting of special relativity, for which only the general Hodge decomposition may be applied, of which Helmholtz is a special case, but even this has nothing to do with it.

This EM field acts on the four-velocity, reproducing the Lorentz force by

$$ \frac{\mathrm{d}p}{\mathrm{d}t} = q F(u)$$

where $u$ is the four-velocity, and $(F(u))_\mu = F_{\mu\nu}u^\nu$.

Let me try this more clearly than the other answers, which aren't wrong. You ask:

So, can someone please elaborate what this EM field is with respect to $\vec E$ and $\vec B$ in the context of Helmholtz decomposition?

There is no "EM field in the context of Helmholtz decomposition". Helmholtz just says that every vector field $\vec V$ is decomposable as curl and gradient of two other fields, i.e.

$$\vec V = \vec \nabla \phi + \vec \nabla \times \vec A $$

You can do this for the electric or the magnetic field, of course, but this isn't particularly enlightening as to the nature of "the EM field". A field should behave nicely under transformations, and special relativity with its action on the electric and magnetic fields shows us that we should not add them together, but seek a quantity that transforms nicely under Lorentz transformations instead:

"The electromagnetic field" is equivalently the gauge four-potential $A$ (consisting of the scalar electrostatic potential in the temporal and the magnetic vector potential in the spatial entries) or its derivative, the field strength tensor $F = \mathrm{d}A$. Electric and magnetic fields become part of the tensor as \begin{align} F^{0i} & = E^i \\ F^{ij} & = \sum_k\epsilon^{ijk}B^k \end{align} This is "the EM field", but it has nothing to do with Helmholtz decomposition, since electromagnetism is properly looked at in the four-dimensional setting of special relativity, for which only the general Hodge decomposition may be applied, of which Helmholtz is a special case, but even this has nothing to do with it.

This EM field acts on the four-velocity, reproducing the Lorentz force by

$$ \frac{\mathrm{d}p}{\mathrm{d}t} = q F(u)$$

where $u$ is the four-velocity, and $(F(u))_\mu = F_{\mu\nu}u^\nu$.