This is where it's a good time to "converse with the math". Let's look at the equation:
$$\mathcal{E} = \oint_\gamma \mathbf{E} \cdot d\mathbf{l}$$
which in this case equals $\mathcal{E}_\mathrm{ind}$. This is based on the work formula:
$$W = \int_\gamma \mathbf{F} \cdot d\mathbf{r}$$
Thus, what your question is, is essentially, asking "how can you have negative work". Looking at the above equation and recalling how a dot product behaves, there's only one way: $\mathbf{F}$ and the element of displacement $d\mathbf{r}$ must be aimed at cross purposes with each other. In the case of the integral for $\mathcal{E}_\mathrm{ind}$, the same goes only with $\mathbf{E}$ and $d\mathbf{l}$.
Hence, the answer to your question is: when $\mathbf{E}$ points opposite to $d\mathbf{l}$.
But what does that mean? Well, the key here is that we have to think a little more closely about the work formula. I believe what you are imagining it means is "the work done by the force as the force pulls the particle along with it". It actually is more general - in a work integral, the particle can be moved in any direction, including against the force. Of course, to make that motion happen in real life, you need to supply a source of contrary force, but that doesn't change the maths. This is why, say, in a more elementary example, you can talk of "negative work" done by the gravitational force when you lift an object off the floor.
(Why is it defined that way? Well, for one, because we often can't solve for the "real" trajectory the particle follows! If we stipulated that as a precondition, it would make work an extremely non-trivial concept!)
Where your mistake lies in, then, is in missing that. The displacement around the wire $d\mathbf{l}$ that we use to describe emf is not the displacement that necessarily occurs in reality to a real positive charge (after all, in many applications we aren't "really" dealing with positive charges anyways!).
Rather, it is a hypothetical one where we imagine that we grab a charge and move it around all the way through the circuit in a specific, fixed direction, and ask what the work - whether positive or negative - done by the electric force for that movement is. If we ask about the work done in the actual movements of charges, we will get a different answer, and yes, this one will always be positive, at least provided we don't get into looking at the situation in too-fine detail.
From an intuitive point of view, if you want to figure when the force is doing "positive" work and when it's doing "negative" work, imagine that you can feel the electric force tugging on the positive charge in your hand as you move it through the circuit. When you feel it helping you, i.e. the tug is with the motion of your hand, at that moment (i.e. that small increment $d\mathbf{l}$) the electric force is doing positive work, and you are doing negative work (to retard the charge, if you were to try and not naturally speed up your hand as you'd likely tend to, of course). When you feel it is fighting you, i.e. the tug is against the motion of your hand, the electric force is doing negative work, and you are doing positive work (to help it against the contrary pull). Total negative work, and hence negative emf, will be if, in moving the charge, you had to fight more often than flow.
in the article linked above it is stated that (3) is the work (per unit charge) done by the Lorentz force
This is strictly speaking incorrect, because the integrand in the formula (3) is not the Lorentz force formula. In the Lorentz force formula, there is velocity of a charged particle, but in the formula (3), the symbol $\mathbf v$ is velocity of an element of wire, which is different from velocity of the mobile charged particles making up the current. Formula (3) is a general formula for total EMF due to both induced electric field, and due to motional EMF due to motion of conductor in magnetic field. The motional EMF in magnetic field is not easily justified as obvious consequence of the Lorentz force formula. As you pointed out, external magnetic force $q_a\mathbf u_a \times \mathbf B$ ($\mathbf u_a$ being velocity of the mobile charged particle $a$) can't do work on the particle $a$ and thus can't contribute to EMF acting on the current.
It takes more steps involving distinguishing forces acting on mobile charges due to a) external field b) internal forces due to wire. It turns out it is the internal forces due to wire lattice and charge distribution on the wire that together are the origin of the elementary forces acting on the mobile charges producing the motional EMF. It just turns out that this result can be expressed as $\mathbf v\times \mathbf B$ where $\mathbf v$ is velocity of the wire, not that of the mobile charge.
Question: What does (3) actually represent? i.e. what actually is the physical meaning of the electromotive force? Is it linked to the work done to move/deform the circuit? If so, can this be shown mathematically?
It is the work done on current in the closed path, per unit current per unit time, at any given instant. It's use is in Kirchhoff's second circuital law, which states sum of all emfs acting on current in a circuit equals sum of terms
$$
R_kI_k
$$
where $R_k$ is resistance of $k$-th member making up the closed conductive path and $I_k$ is current flowing through it. For a single conductive closed path (loop of wire), there is only one current, so we can simplify and say that
$$
\text{total emf} = RI
$$
where $R$ is total resistance of the circuit.
Best Answer
You're confusing induced EMF with total EMF, which is sum of all contributions to EMF, including the induced EMF(due to induced electric field, described by the Faraday law) and the battery EMF (due to electrochemical processes in the battery, not described by the Faraday law).
Net EMF due to both induced field and battery can be written as
$$ \mathscr{E}_{net} = \mathscr{E}_{induced} + \mathscr{E}_{battery} $$ $$ \mathscr{E}_{net} = \oint \mathbf E_{induced} \cdot d\mathbf s + \mathscr{E}_{battery}. $$ $$ \mathscr{E}_{net} = -\frac{d}{dt}\int \mathbf B \cdot d\mathbf S + \mathscr{E}_{battery}. $$
If resistance of the circuit is $R$, then Kirchhoff's second circuital law predicts that current due to both sources of EMF will be
$$ I = \frac{\mathscr{E}_{net}}{R}. $$