(a) The total emf in one coil (2) is
$$\mathscr E_2 = -\frac{d\phi_\text{linked with 2}}{dt}=-L_2\frac{dI_2}{dt}-M\frac{dI_1}{dt}.$$
We can, if we wish, regard this as the sum of two emfs.
(b) You are quite right that the changing current induced in 2 produces changing flux some of which is linked with coil 1 and affects the current in it, which in turn affects the emf in coil 2, according to the equation above. But this is not double-counting: we simply have to accept that $I_1$ and $I_2$ are not independent of each other.
I finally have some time to type this up, and show you how and why KVL is just the wrong way to deal with this.
Faraday's Law is, with minus sign as the Lenz's Law,
$$\tag1\oint_{\partial S}\vec E\cdot\vec{\mathrm d\ell}
=-\frac{\mathrm d\ }{\mathrm dt}\iint_S\vec B\cdot\vec{\mathrm d^2A}$$
Note that the LHS is an integration over the electric field, and the RHS is an integration over the magnetic field. This is a statement of a relationship between two different phenomena, and should be kept as such.
Now, let us apply Faraday's Law to the middle circuit. Integrating clockwise starting from the switch, the first contribution is the electric field inside the battery, and we are traversing from the negative terminal to the positive terminal. A battery is a chemical soup that maintains the EMF voltage difference between its terminals, but for our current integration purposes, it is the same as if a capacitor. We are going against the electric field, and so it gives us $-\mathscr E$ as the contribution. After the battery, we are going from positive terminal to negative via the usual route, and the current $I_2$ that is drawn, is in the correct direction, i.e. $I_2>0$. Going down the resistor, the electric field integration will be $+I_2R_2$. Finally, we integrate the negligible electric field along the loops of the solenoid, and return to the switch. The inductance is a magnetic effect, and it appears on the RHS. I will not pretend to know the directions involved; we will fix them later.
$$\tag2\therefore\qquad-\mathscr E+I_2R_2=-L_2\dot{I_2}+M_{21}\dot{I_1}+M_{23}\dot{I_3}$$
Rearranging the terms, we have $L_2\dot{I_2}+I_2R_2=\mathscr E+M_{21}\dot{I_1}+M_{23}\dot{I_3}$; the LHS of this equation reproduces the usual expression for LRC circuits and so must be correct, thereby telling us that the sign choice is correct. Now, note that KVL, strictly speaking, should not have any magnetic terms appearing. It is only via the brutal hacking away at Faraday's Law, rearranging terms haphazardly, that we would have KVL appearing to work. Such hacking is very easily going to make up confusing experimental results. For example, if you put in a voltmeter inside this circuit, it is possible to engineer things so that the voltmeter reading can differ depending upon the location of the voltmeters and the wires connecting them. Here is a video of simultaneous measurement of the same two points, with the oscilloscopes put on opposite sides. And it is very annoying that YouTube is not automatically recommending the part two of that same video.
Instead of working out what the signs of the mutual inductances should be right away, let us consider the easier circuits on the side. We know that $I_1$ is definitely going to be increasing from zero. This means that a new magnetic $\vec B$ field will be created in the solenoid $L_2$ with North pole pointing towards $L_1$. This means that Lenz's Law says that the solenoid $L_1$ will be creating a $\vec B$ field that with North pole pointing towards $L_2$, attempting to negate what is being thrown at it. Thus, we can write down that the left loop must be having
$$\tag3I_1R_1=-L_1\dot{I_1}+M_{12}\dot{I_2}$$
where we are finally, for the first time, fully aware that this mutual inductance term is chosen with the correct sign. The right loop is not even a loop, so we can only guess, using similar reasoning, that it should be
$$\tag40?=-L_3\dot{I_3}+M_{32}\dot{I_2}$$
unless we know what is to be connected between those terminals, replacing the $0?$.
Now, we try to fix the signs on Equation (2). If we imagine taking away the battery, and only consider increasing $I_1$ and $I_3$ independently from zero, then it is clear that the signs there are chosen correctly. Of course, I am assuming that all of $M_{12},M_{23},L_1,L_2,L_3,\mathscr E>0$ and we know by definition that $M_{21}=M_{12}\ \wedge\ M_{32}=M_{23}$; all resistances are positive, or else they won't even make sense. Mind you, the blue arrows defined the positive directions of all the currents, and we now know that all three of them are going to increase from zero when the switch is initially closed.
Note that I have given you not just the correct solution, but that the scheme is completely general: the only way to make sure all the signs are correct, is to actually consider the electric fields and magnetic fields in a theoretically coherent manner. Otherwise, it will have too many possible avenues for making a mistake.
Best Answer
Unless they are obvious from context, the polarities of coupled inductors should normally be indicated by dots on the "positive" terminal of each inductor, like in the diagram below (image credit). This means that if inductor currents are defined as entering the positive terminals, the mutual inductance will be positive.
In your case, the polarities are ambiguous. It sounds like they just want to take the currents as being defined in the diagram and take $M > 0$, but know that this isn't normally a given and ideally in the real world you would be given more information.