Now that I know what to look for, it seems some of the pieces are starting to fall into place. Part 1 was about Levels $n=1,2,8,9$ of Ramanujan-Sato series. Part 2 below involves Levels $n=4,6,10$. (But I'm missing the odd prime levels $n=3,5,7,11$.)
I. Question
Given binomial $\binom{n}{k}$ and any $A,B,C$. If the series converges, is it true that,
$$\frac{1}{C^{3/2}}\sum_{k=0}^\infty s_1(k)\frac{A\,k+B}{C^{k}}=\\\frac{1}{(C+4r)^{3/2}}\sum_{k=0}^\infty s_2(k) \frac{A\,k+B-Dr}{(C+4r)^k}$$
given the sequences,
$$s_1(k) = \tbinom{2k}{k}^{2-p}\sum_{j=0}^k\tbinom{k}{j}^{2+p}\\ \color{red}{s_2(k)} = \tbinom{2k}{k}\sum_{j=0}^k r^{k-j}\tbinom{k}{j} \tbinom{2j}{j}^{1-p}\sum_{m=0}^j\tbinom{j}{m}^{2+p}$$
where $D = \frac{2A-4B}C$, and for any real $r$ such that $\color{brown}{C(C+4r)>0}$?
(Again, if the LHS is a formula for a well-known constant, then the RHS guarantees infinitely many such formulas.)
II. Ramanujan-Sato level 4
If $p=0$, then $s_1(k) = \tbinom{2k}{k}^{2}\sum_{j=0}^k\tbinom{k}{j}^{2}=\tbinom{2k}{k}^3$. For example, starting with,
$$\frac1{\pi} =\frac{-2^{12}\sqrt{-1}}{(-2^9)^{3/2}} \sum_{k=0}^\infty\tbinom{2k}{k}^3\,\frac{6k+1}{(-2^9)^k}$$
Since $D = \frac{2A-4B}C = -1/2^6$, then,
$$\frac1{\pi} =\frac{-2^{12}\sqrt{-1}}{(-2^9+4r)^{3/2}} \sum_{k=0}^\infty s_2(k)\frac{6k+1+r/2^6}{(-2^9+4r)^k}$$
with the sequence $s_2(k)$ as defined in the question.
III. Ramanujan-Sato level 6
If $p=1$, then $s_1(k) = \tbinom{2k}{k}\sum_{j=0}^k\tbinom{k}{j}^{3}$ which is OEIS sequence A181418. For example, from,
$$\frac1{\pi}=\frac{192\sqrt3}{(2\cdot140^2)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}k\sum_{j=0}^k\tbinom{k}{j}^3\,\frac{140(561k+53)}{(2\cdot140^2)^k}$$
then,
$$\frac1{\pi}=\frac{192\sqrt3}{(2\cdot140^2+4r)^{3/2}}\sum_{k=0}^\infty s_2(k)\frac{140(561k+53)-13r/4}{(2\cdot140^2+4r)^k}$$
IV. Ramanujan-Sato level 10
If $p=2$, then $s_1(k) =\sum_{j=0}^k\tbinom{k}{j}^4$. For example, from
$$\frac1{\pi}=\frac{16\sqrt{\tfrac5{19}}}{(76^2)^{3/2}}\sum_{k=0}^\infty \sum_{j=0}^k\tbinom{k}{j}^4\,\frac{19^2(408k+47)}{(76^2)^k}$$
then, $$\frac1{\pi}=\frac{16\sqrt{\tfrac5{19}}}{(76^2+4r)^{3/2}}\sum_{k=0}^\infty s_2(k)\frac{19^2(408k+47)-157r/4}{(76^2+4r)^k}$$
So, like in Part 1, is the answer to the question a "yes"?
