The reason we put dots on indices is to signify that a transformation of the $\alpha$ index is independent of a transformation on the $\dot{\alpha}$ index. If they were related by a similarity transformation we wouldn't need to use dots, so this dot stuff is completely irrelevant in a $\mathrm{SU}(2)$ discussion
$\psi^1 \psi^{*1} + \psi^2 \psi^{*2}$ is invariant under $\mathrm{SU}(2)$ because as I've said multiple times now it's just $\psi^{\dagger} \psi$ and $\psi'^{\dagger} \psi' = \psi^{\dagger} U^{\dagger} U \psi = \psi^{\dagger} \psi$
You didn't transform the delta properly, the minus sign is just a mistake, and there are absolutely no dotted indices. The whole thing I just posted is summarized by $\psi'^{\dagger} \psi' = \psi^{\dagger} U^{\dagger} U \psi = \psi^{\dagger} \psi$, but I did it the way you typed it which amounts to writing: $$\psi'^{\dagger} I \psi' = \psi^{\dagger} U^{\dagger} [U^{\dagger} I U] U \psi = \psi^{\dagger} U^{\dagger} U \psi = \psi^{\dagger} \psi$$ seen this way, it's absurd to transform the (invariant) $I$ which you then just cancel but the confusing indices make it look like you have to do it
You transformed $\delta$ wrong again, if you write $F F^{\dagger} = I$ and $F^{\dagger} F = I$ out in indices, and also write $F I F^{\dagger} = I$ and $F^{\dagger} I F = I$ out in indices, you will see this
That's very hard to read and the equal signs are missing, look up the latex code for aligning equations
Your index convention will break, it's unavoidable, you're making mistakes by not using this, and there should be absolutely no minus signs anywhere
L&L wrote $\psi^{1} \psi^{*1} + \psi^2 \psi^{*2}$, this is trivially equal to $\delta_{\alpha \beta} \psi^{\alpha} \psi^{*\beta}$, you are making mistakes with this formal space nonsense, it's a trivial equality
No, $\psi^{\alpha}$ is a complex number, $\psi^{*\alpha}$ is another complex number, I multiplied two complex numbers $\psi^1$ and $\psi^{*1}$, did the same with $\psi^2$ and $\psi^{*2}$, then just added the result of this trivial multiplication when I wrote $$\delta_{\alpha \beta} \psi^{\alpha} \psi^{*\beta} = \psi^{1} \psi^{*1} + \psi^2 \psi^{*2}$$
A given $F \in \mathrm{SU}(2)$ must satisfy $F F^{\dagger} = I$, in terms of indices this reads as $F^{\alpha}_{\ \ \ \ \kappa} (F^{\dagger})^{\kappa}_{\ \ \ \beta} = F^{\alpha}_{\ \ \ \ \kappa} F^{* \beta}_{\ \ \ \ \ \ \kappa} = \delta^{\alpha \beta}$. You are literally denying the most basic definition of $\mathrm{SU}(2)$ because you don't want to write out $\dagger$ as a conjugate transpose and it's leading you so far away that you're finding minus signs...
Why do you keep summing over dropped indices like that. You have to be careful because if you simultaneously raise and lower the same index, you get a minus sign i.e. $S^{\alpha} S_\alpha = -S_\alpha S^{\alpha}$
$F^\alpha{}_\kappa (F^\dagger)^\kappa{}_\beta = \delta^\alpha{}_\beta$ If you raise the index you need another epsilon tensor.
But this delta is actually different from the lowered delta you used in your definition which is actually the epsilon tensor
A matrix $M_{ab}$ satisfies $M_{ab}^T = M_{ba}$, that is more fundamental than any up or down notation
When you transformed the $\delta$ you did it wrong, you wrote $\delta_{\alpha \beta} = F_{\alpha}^{\ \ \gamma} F_{\beta}^{\ \ \delta} \delta_{\gamma \delta}$ with no explanation, what you're supposed to be doing is writing out $F^{\dagger} I F = I$ in indices, but what you did is write $F I F = I$
On page 118 he writes $M_{\alpha}^{\ \ \beta} = (M^{\dagger})_{\alpha}^{\ \ \beta} = (M_{\beta}^{\ \ \alpha})^*$ 'breaking' the index notation
When you transformed $\psi^{* \alpha}$ you incorrectly wrote a $F^{\alpha}_{\ \ \phi}$ when it should be $F^{\phi}_{\ \ \alpha}$ due to the transpose which is even what he does too
When he took the transpose he put the $\beta$ down and on the left, while you kept it up when moving it to the left, this actually breaks the definition of the matrix unless you use $\epsilon$'s which you didn't mention
Ok forget about any raising and lowering of indices, that's an additional concept. Let $\chi^{\alpha}$ be a column vector and $\psi_{\alpha}$ be a row vector. Clearly $(\psi^{\alpha})^T = \psi_{\alpha}$ just says the transpose of a column vector is a row vector. If you have a matrix $M^{\alpha}_{\ \ \beta}$ the indices are placed where they are so that $\psi_{\alpha} M^{\alpha}_{ \ \ \beta} \chi^{\beta}$ has a nice Einstein summation pattern.
It's transpose is now $(M^{\alpha}_{\ \ \beta})^T = M^{\beta}_{\ \ \alpha}$ which ensures $(\psi_{\alpha} M^{\alpha}_{ \ \ \beta} \chi^{\beta})^T = \chi_{\beta} M^{\beta}_{\ \ \alpha} \psi^{\alpha}$ is consistent. It can't be $(M^{\alpha}_{\ \ \beta})^T = M_{\beta}^{\ \ \alpha}$, this breaks the transpose of $(\psi_{\alpha} M^{\alpha}_{ \ \ \beta} \chi^{\beta})$
and what you really you want is code that doesn't need a lot of comments because it is well-structured and has appropriately-named variables :P
@SirCumference I never know which option is worse: Someone wrote such a comment genuinely thinking it would be helpful, or they just wrote it because someone told them to "write comments"
I once had to dive into old code apparently written by someone new to the concept of objects and references where every object reference was just named some variant of "ref" with various useless prefixes/suffixes
