Not sure what you did wrong, but I suspect that you ignored something in your casework. Also, the total number of cases should be $n^{n+1}$, not $n^{n+2}$. Here is an example of a much simpler approach.
First, calculate the number of ways there can be two empty bins. The number of ways for the two empty bins to be chosen is
$$_{n}C_2=\frac{n(n-1)}{2}$$
then, since there must be exactly two empty bins, we must place one ball in each of the remaining $n-1$ bins so that none of them are empty. We then have one last ball with $n-1$ choices. Thus the number of ways to place the balls and have exactly two empty bins is
$$\frac{n(n-1)}{2}*(n-1)$$
$$\frac{n(n-1)^2}{2}$$
And so the probability is
$$\frac{\frac{n(n-1)^2}{2}}{n^{n+1}}$$
$$\frac{(n-1)^2}{2n^{n}}$$
EDIT: This is incorrect. By saying that the number of ways to distribute the balls is $n^{n+1}$, I imply that the balls are distinct objects; however, if they are distinct, then the number of ways to place one ball in each bin other than the two designated empty bins is $(n-1)!$, and so the number of ways to place the balls and have two empty bins is
$$\frac{n(n-1)}{2}*(n-1)!*(n-1)=\frac{n(n-1)^2(n-1)!}{2}$$
and so the probability is
$$\frac{\frac{n(n-1)^2(n-1)!}{2}}{n^{n+1}}$$
$$\frac{(n-1)^2(n-1)!}{2n^n}$$
The placement of the non-red balls is irrelevant...the problem is decided by the placement of the red ones alone.
To do the first part, it is easier to work from the complimentary event. The probability that a specified red ball goes to a bin other than the second is $\frac 34$. Thus the answer is $$1-\left( \frac 34\right)^2=\boxed {\frac 7{16}}$$
To do the second one, it might be helpful to simply list the possible placements of the two red balls. Given that the second bin must contain a red ball, there are only four cases: $$(1,1,0,0)\quad (0,2,0,0)\quad (0,1,1,0)\quad (0,1,0,1)$$
Where writing, say, $(1,1,0,0)$ means that the first two bins each have one red ball and the second two each have none.
Note that the probability of getting $(0,2,0,0)$ is $\frac 1{16}$ since we must have both red balls going to the second bin. The probability of getting, say, $(1,1,0,0)$ is $\frac 18$ since we can get this configuration in two ways (either the first red ball goes in the first bin and the second red ball goes in the second bin, or the first red ball goes in the second bin and the second red ball goes in the first bin). This gives another way to see $\frac 7{16}$ as the result for the first problem, since $$\frac 18+\frac 1{16}+\frac 18+\frac 18=\boxed {\frac 7{16}}$$
We see from this that the conditional probability that the first bin has a red ball (conditioned on the fact that the second bin also has one) is $$\frac {1/8}{7/16}=\boxed {\frac 27}$$
Side note: if you prefer to work with equi-probable scenarios (not a bad idea) then you need to indicate the placement of $r_1, r_2$ (the two red balls) separately. Thus the scenario $(1,1,0,0)$, say, becomes two scenarios, $(r_1,r_2,0,0)$ and $(r_2,r_1,0,0)$. You wind up with seven scenarios in which the second bin contains at least one red ball, each of which having probability $\frac 1{16}$.
Best Answer
This seems computationally very difficult to me. You might do better with simulation.
It $C\ge m$ then the constraint has no effect, and we are just looking for the mean and variance of the number of balls in bin $j.$ This has a binomial distribution, and the answer is well known.
When ${m\over k}\leq C< M,$ the situation is much more complicated, because the result depends on the sequence in which the balls were thrown into the bins. Before any bin fills up, a ball has probability $1/k$ of landing in bin $j.$ After some other ball fills up, it has probability $1/(k-1)$ of landing in bin $j$. The situation is even more complicated if it's possible for more than one ball to fill up.
Consider the case $k=3, j=1.$ Let $P(x_1,x_2,x_3)$ be the probablity that at some point there are $x_i$ balls in bin $i$ for $i=1,2,3.$ To compute the expectation, we have to sum up terms of the form $aP(a,b,c)$ where $a+b+c=m.$ If $\max\{a,b,c\}<C,$ we have simply $$P(a,b,c)={m\choose a,b,c}3^{-n}\tag{1}$$ But suppose $c=C.$ Then $$P(a,b,c)=P(a,b,C)=\frac12P(a-1,b,C)+\frac12P(a,b-1,C)+\frac13P(a,b,C-1)\tag{2}$$ taking account of all bins the last ball might have been thrown into.
The last term on the right-hand side of $(2)$ can be computed directly from $(1),$ but the first two have to be computed recursively from $(2)$. It won't take many bins, or many balls, before this computation becomes unwieldy, if the capacity constraint is effective.
There are so many possible sequences of throws that memoization is unlikely to be effective, so far as I can see. I've been tying to think of ways to simplify the calculations, but so far, I don't have a glimmer.