Without going into Haag's theorem, I will briefly comment on why we use the interaction picture.
For your first question
What we want to calculate is n-point functions:
$$\langle 0|T\phi_H(x_1)\cdots\phi_H(x_n)|0\rangle,$$
where $\phi_H$ is the field operator of Heisenberg representation and satisfies the EoM of the interacting theory. To calculate these n-point function, it is useful to define the partition function:
$$Z=\langle0|T\exp\Big(i\int d^4x\ j(x)\phi_H(x)\Big)|0\rangle,$$
$$\langle 0|T\phi_H(x_1)\cdots\phi_H(x_n)|0\rangle:=\frac{\delta^n Z[j]}{i\delta j(x_1)\cdots i\delta j(x_n)}.$$
So what we should do is replaced to calculate the partition function $Z$. Also, we are interested in perturbative calculations, we must separate interacting terms from free terms so that we can use a perturbative expansion. However it is quite nontrivial how to treat $Z$ perturbatively because the interaction term and the free term are not separated explicitly, we do not know how to handle them in a perturbative calculation.
Therefore, we split the Hamiltonian as follows,
$$H=H_0+H_{\mathrm{int}}$$
with $H_0$ carrying the time evolution of the field operators and $H_{\mathrm{int}}$ carrying the time evolution of the states. Under such a division, the operator in the Heisenberg representation and the operator in the interaction representation have the following relationship:
$$\phi_H(x)=U^\dagger(t,-\infty)\phi_I(t,\mathbf{x})U(t,-\infty),$$
$$U(t_1,t_2):=T\exp\Big(-i\int_{t_1}^{t_2}dt\ H_{\mathrm{int}}\Big),$$
$$U^\dagger(t_1,t_2)=U(t_2,t_1),$$
$$U(t_1,t_2)U(t_2,t_3)=U(t_1,t_3),$$
where we assume that $\phi_I$ and $\phi_H$ coincide at the far past $t\to -\infty$. (This coincide to choose a reference time to be $t_0=-\infty$.) This is natural assumption in the scattering theory because particles will be well-separated at far past.
Using the fierst equation above, we can rewrite the n-point functions as
$$\langle 0|T\phi_H(x_1)\cdots\phi_H(x_n)|0\rangle=U(-\infty,+\infty) \langle 0|T\phi_I(x_1)\cdots\phi_I(x_n) \exp\Big(i\int_{-\infty}^{\infty}dt\int d^3 x\ \mathcal{L}_{\mathrm{int}}(\phi_I(x))\Big) |0\rangle,$$
where we define the Lagrangian density as follows $\mathcal{L}_{\mathrm{int}}:=-\mathcal{H}_{\mathrm{int}}. $
Thus we don’t need to use the complicated expression
$$Z=\langle0|T\exp\Big(i\int d^4x\ j(x)\phi_H(x)\Big)|0\rangle,$$
but all we have to do is to evaluate
$$Z= \langle 0|U(-\infty,+\infty) T\exp\Big(i\int d^4 x\ (j(x)\phi_I(x)+\mathcal{L}_{\mathrm{int}}(\phi_I(x)))\Big) |0\rangle$$
$$=\exp\Big(i\int d^4x \mathcal{L}_{\mathrm{int}}(\frac{\delta}{i\delta j(x)}) \Big) \langle0|T\exp\Big(i\int d^4x\ j(x)\phi_I(x)\Big)|0\rangle$$
This expression is useful because it nicely separates the free and interacting parts. This is the reason for introducing the interaction picture. Also, since $\phi_I$ satisfies the free-field equation of motion, this allows $\langle0|T\exp\Big(i\int d^4x\ j(x)\phi_I(x)\Big)|0\rangle$ to be rewritten by taking the appropriate normal order as
$$\langle0|T:\exp\Big(i\int d^4x\ j(x)\phi_I(x)\Big):|0\rangle=\exp\Big(-\frac{1}{2}\int d^4x d^4y \ j(x)j(y)\Delta_0(x-y)\Big),$$
where $\Delta_0(x-y)$ is a free propagator.
This allows us to understand that the derivative with respect to $j$ contained in the interaction term acts one after the other on $\langle0|T\exp\Big(i\int d^4x\ j(x)\phi_I(x)\Big)|0\rangle$, thereby producing a free-field propagator. This is another simplification that arises because $\phi_I$ satisfies the free-field equation of motion.
In other words, the interaction picture is one prescription for successfully separating the free-field contribution from the interaction contribution for perturbation calculations.
For your second question
I simply don't understand what you are saying. Perhaps you are confused between the various time evolution operators, so I suggest you rethink it yourself, referring to the examples I gave above, etc.
