Rutherford modeled the atom as an extremely compact positive nucleus surrounded by a uniform ball of negative charge the "size" of the atom. He included the effect of scattering by the electrons under the assumption that they acted like such a diffuse cloud of negative charge, and showed that such a cloud had negligible probability of scattering the alpha particles by more than a degree. That's not because it lacks mass, but because it is diffuse, or rather, composed of many small charges rather than one compact point of high charge, so the alpha particle trajectory would be composed of multiple small deflections ("deflexions" in his spelling!) from the negative charges rather than a single large one. So yes, the electrons could scatter the alpha particles, but for a diffuse electron cloud the effect is negligible. (Actually, as far as I can tell the data would also be consistent with a compact negative nucleus surrounded by a diffuse positive cloud. How Rutherford justified the conclusion that the nucleus was the positive part is mysterious to me.)
Edit: From Rutherford, Philosophical Magazine 21, 669-688 (1911), penultimate paragraph: "The deductions from the theory so far considered are independent of the sign of the central charge, and it has not so far been found possible to obtain definite evidence to determine whether it be positive or negative." He goes on to explain why a positive nucleus might make more sense in terms of $\beta$ absorption and $\alpha$ emission (e.g. why $\alpha$ particles are emitted with such high velocity).
From the same article: Rutherford explains that J. J. Thomson's model of distributed positive and negative charges involves multiple small scattering events, and predicts only small overall "deflexions" (pp 669-670).
After laying out his own model of the compact central charge surrounded by a uniformly dense negative charge, he writes (pp. 671-2), "Since R [the radius of the ball of negative charge] is supposed to be of the order of the radius of the atom, viz. $10^{-8}$ cm., it is obvious that the $\alpha$ particle before being turned back penetrates so close to the central charge, that the field due to the uniform distribution of negative electricity may be neglected. In general, a simple calculation shows that for all deflexions greater than a degree, we may without sensible error suppose the deflexion due to the field of the central charge alone. Possible single deviations due to the negative electricity, if distributed in the form of corpuscles, are not taken into account at this stage of the theory. It will be shown later that its effect is in general small compared with that due to the central field."
He gets around to that discussion near the end, which may be most relevant to your question (p. 686): "In comparing the theory outlined in this paper with the experimental results, it has been supposed that the atom consists of a central charge supposed concentrated at a point, and that the large single deflexions of the $\alpha$ and $\beta$ particles are mainly due to their passage through the strong central field. The effect of the equal and opposite compensating charge supposed distributed uniformly throughout a sphere has been neglected. Some of the evidence in support of these assumptions will now be briefly considered. For concreteness, consider the passage of a high speed $\alpha$ particle through an atom having a positive central charge Ne, and surrounded by a compensating charge of N electrons. Remembering that the mass, momentum, and kinetic energy of the $\alpha$ particle are very large compared with the corresponding values for an electron in rapid motion, it does not seem possible from dynamic considerations that an $\alpha$ particle can be deflected through a large angle by a close approach to an electron, even if the latter be in rapid motion and constrained by strong electrical forces. It seems reasonable to suppose that the chance of single deflexions through a large angle due to this cause, if not zero, must be exceedingly small compared with that due to the central charge."
BTW, I highly recommend reading the original article. Once you decipher the somewhat archaic notation, it's extremely readable.
For the first picture, you are right. The force on the $\alpha^{++}$ particle is twice that on the $\beta^-$ particle, but also the velocity of the $\alpha^{++}$ is much smaller, so it's easier to change direction.
In the second case, the centripetal force needed is much higher for the particle with larger mass,
$$q\vec v\times\vec B=\frac{mv^2}{r}$$
so $r$ is much larger due to the large $m$, and double charge does not affect it significantly.
Best Answer
The usual derivation of the differential scattering cross section makes the assumption that the mass of the target nucleus is much greater than that of the incoming alpha particle. This is saying that the nucleus does not recoil when it interacts with alpha particle. Better still assume that the $\csc^4$ formula derivation is done in the centre of mass frame of the nucleus and alpha.
Since observations are made in the laboratory frame a correction has to be applied to the $\csc^4$ formula and that correction does depend on the mass of the nucleus in relation to the mass of the alpha and so changing the number of neutrons whilst keeping the number of protons the same does change the angle of deflection.
The smaller the nucleus the greater the correction term when the number of neutrons changes.
There are many Internet sites and textbooks which derive the correction term and here is one with the centre of mass derivation shown on the previous page.