Yes, Abel's argument, particularly in the original 1824 paper, is incredibly condensed! I believe that this is partly because he was printing the proof at his own expense as he wanted it as a 'calling card' for other mathematicians. I'll try to answer your questions here. Out of interest, have you posted questions on the rest of the proof as you indicated you might?
Question 1
In fact Abel initially says that any root of the given quintic can be put in the form:
$$y=p+p_1R^{\frac{1}{m}}+p_2R^{\frac{2}{m}}+\ldots +p_{m-1}R^{\frac{m-1}{m}}$$
which is not quite the same as what you say, i.e. this goes up an arbitrary $m$. You indicate that you're happy with how he gets to the form $\frac{F}{G}$ so I'll only briefly mention that he assumes that if one has an expression with many nested fractions, i.e. fractions within fractions, then by repeatedly multiplying through by denominators we can eliminate virtually all of them until we reach something of the form $\frac{F}{G}$. Both numerator and denominator are likely to contain radicals, i.e. if we choose one of the 'outer radicals', i.e. not nested inside any other, and call this $S^\frac1m$, then the denominator will be of the form $u+u_1 S^\frac1m+u_2 S^\frac2m+\ldots$, i.e. in terms of $S^\frac1m$ and possibly some of its powers.
An important point to answer your question 2 is that if is ever possible to write this radical as a rational function of the other terms in the function then this is done, removing that particular root and choosing another.
Having done this, Abel shows how it is possible to 'rationalise the denominator' to eliminate the chosen radical from the denominator in similar way to the to the usual trick of multiplying a denominator of the form $a+\sqrt{b}$ by $a-\sqrt{b}$. The general method if the radical is an $m$-th root, i.e. for a denominator of the form $U=u+u_1 S^\frac1m+u_2 S^\frac2m+\ldots+u_{b} S^{\frac{b}{m}}$, is to multiply top and bottom by the functions $U_1,U_2, \ldots U_{m-1}$ formed by replacing $S^\frac1m$ in $U$ with $\alpha S^\frac1m, \alpha^2 S^\frac1m, \ldots \alpha^{m-1}S^\frac1m$ where $\alpha$ is an $m^\textrm{th}$ root of unity other than 1. Terms such as $S^\frac2m$ therefore become $(\alpha S^\frac1m)^2$ in $U_1$ and $(\alpha^2 p^\frac1m)^2$ in $U_2$ etc. When the denominator is multiplied out and simplified using the fact that $1+\alpha+\dots+\alpha^{m-1}=0$, all fractional powers of $S$ cancel.
Abel seems to indicate that the technique was standard in his time, but as it's not so common now, an example may help, namely that to rationalise a denominator such as $3+\sqrt[3]{2}+(\sqrt[3]{2})^2$, one can multiply by $3+\alpha \sqrt[3]{2}+(\alpha\sqrt[3]{2})^2$ and $3+\alpha^2 \sqrt[3]{2}+(\alpha^2\sqrt[3]{2})^2$ where $\alpha$ is a cube root of $1$ other than $1$, giving:
$$(3+\sqrt[3]{2}+(\sqrt[3]{2})^2)(3+\alpha \sqrt[3]{2}+(\alpha\sqrt[3]{2})^2)(3+\alpha^2 \sqrt[3]{2}+(\alpha^2\sqrt[3]{2})^2)$$
It takes some working through, but using $1+\alpha+\alpha^2=0$ eventually shows that this equals 15, i.e. with no $\sqrt[3]{2}$.
So if we also multiply the numerator of $\frac{F}{G}$ by this expression then we will end up with a fraction where the denominator now only contains integer powers of $S$ and so we can incorporate the denominator into each of the coefficients of $S^\frac1m$, leading to an expression of the form:
$$y=v+v_1 S^\frac1m+v_2 S^\frac2m+\ldots+v_c S^{\frac{c}{m}}$$
The final step is to get rid of any higher powers of $S^\frac1m$ than $m-1$, achieved using the fact that $S^{\frac{m}{m}}=S$. By `cancelling down' any powers of $S^\frac1m$ greater than or equal to $m$ and relabelling the coefficients once again as $q,q_1,\ldots$ we reach the form given above.
Question 2
As you say, Abel then substitues this expression for $y$ back into the original quintic. As we are raising $y$ to various powers, this will results in an expression containing many higher powers of $R^\frac1m$, which still equals 0. Using $R^\frac{m}{m}=R$ any power of $R^\frac1m$ higher than $m-1$ can be reduced to below $m$. Using this and collecting like powers of $R^\frac1m$ together leads to an expression of the form
$$q+q_1 R^\frac1m+q_2 R^\frac2m+q_3 R^\frac3m+\ldots+q_{m-1} R^{\frac{m-1}{m}}=0$$
where $q, q_1, q_2,\ldots q_{m-1}$ are rational functions of $p, p_2, \ldots$ along with the coefficients of the original polynomial.
Abel now proves that all of $q, q_1,q_2 \ldots q_{m-1}$ will turn out to be zero, or in other words, that when multiplying out the expression formed by substituting $y$ into a polynomial and reducing any powers of $R^\frac1m$ higher than $m-1$, all of the various terms containing each power of $R^\frac1m$ will cancel out, i.e. all terms containing $R^\frac1m$ will cancel, as will any containing $R^\frac2m$, $R^\frac3m$ etc.
His overall strategy is to derive a contradiction from assuming that not all of $q, q_1,q_2 \ldots q_{m-1}$ are zero. He notes to this end that if this equation in $R^\frac1m$ is true then $R^\frac1m$ is a root of the polynomial
$$p+q_1 z+q_2 z^2+q_3 z^3+\ldots+q_{m-1} z^{m-1}=0$$
Now, being an $m$-th root, $R^\frac1m$ is also a root of $z^m-R=0$ and there may or may not be other roots common to both equations. Abel says that is be possible to find the polynomial whose roots are all of the roots common to these two equations, i.e. if there are $k$ common roots this means a polynomial of the form
$$r+r_1 z+r_2 z^2+r_3 z^3+\ldots+r_k z^k=0$$
where, crucially, the coefficients $r,r1,r2,\ldots r_k$ are rational function of the $q, q_1, \ldots, q_{m-1}$ and $R$. Abel does not give details on how to do this in either paper, simply saying that it is known that it is possible, but one likely method, well known to Abel, would be to use the Euclidean algorithm to find the highest common factor of the two expressions.
Abel then shows that there is a contradiction with the fact that $R^\frac1m$ cannot be expressed as a rational function of $p, p_2,\ldots p_{m-1}$ by noting that all of these common roots will be of the form $\alpha_\mu z$, where $\alpha$ is an $m^{\textrm{th}}$ root of unity other than 1 and $z=R^\frac1m$ (there is a slight inconsistency in notation here, with $z$ being used both for the variable in the equations as well as one of their roots $R^\frac1m$, for clarify Abel should perhaps write $z'=R^\frac1m$). If there are $k$ roots in common, one of which is $z$ and another is $\alpha z$ then the remainder are $\alpha_1,\ldots,\alpha_{k-2}$. Substituting each of these into the earlier equation give the simultaneous equations:
\begin{equation}
\begin{aligned}
r+r_1 z+r_2 z^2+r_3 z^3+r_k z^k&=0\\
r+\alpha r_1 z+\alpha^2 r_2 z^2+\ldots+\alpha^2 r_k z^k&=0\\
\ldots&\ldots\\
r+\alpha_{k-2} r_1 z+\alpha_{k-2}^2 r_2 z^2+\ldots+\alpha_{k-2}^k r_k z^k&=0\\
\end{aligned}
\end{equation}
These can be treated as $k$ linear simultaneous equations in $z, z^2,\ldots z^k$ and so it is possible to find an expression for $z$ as a rational function of $r,r_1,\ldots r_k$, which, as mentioned above, in turn are rational functions of $q, q_1, \ldots, q_{m-1}$ and $R$, which in turn, again as mentioned above, are polynomial functions of $p, p_2,\ldots$ along with the coefficients of the original polynomial. This contradicts the fact that $R^\frac1m$ cannot be expressed as a rational function of $p, p_2,\ldots p_{m-1}$ and so all of $q, q_1,q_2 \ldots q_{m-1}$ are zero.
Question 3
The previous reasoning has shown that substituting $R^\frac1m$ into the quintic results in an expression which can only equal zero if all of the various powers of $R^\frac1n$ cancel out. As you say, Abel points out next that the same thing will happen if $R^\frac1m$ is replaced in the expression for $y$ by any of $\alpha R\frac1m,\alpha^2 R\frac1m,\ldots \alpha^{m-1} R\frac1m,$ where $\alpha$ is an $m^\textrm{th}$ root of unity other than 1. He simply says that 'it is clear' that is the case, but his reasoning is most likely that if we replace $R^\frac1m$ with $\alpha^i R^\frac1m$ in $y$ and then substitute this new $y$ into the given polynomial then all resulting terms containing a given power of $R^\frac1m$ will contain the same power of $\alpha^i$, e.g. any term containing $R^\frac2m$ will also contain $\alpha^{2i}$. When the powers of $R^\frac1m$ greater than $m-1$ are reduced using $R^\frac{m}{m}=R$, i.e. reducing the power by $m$, the powers of $\alpha$ can also be reduced by $m$ given that $\alpha^m=1$. Each term containing $R\frac1m$ in the expression for $y$ after replacing $R^\frac1m$ with $\alpha^i R\frac1m$ will therefore contain the same power of $\alpha$, meaning that this power of $\alpha$ can be factorised out of the terms. If the terms cancelled before the $\alpha$'s were introduced, i.e. as shown by $q, q_1,q_2 \ldots q_{m-1}$ all being zero, then the terms will also cancel after $R\frac1m$ is replaced by $\alpha^i R\frac1m$, i.e. all of the new expressions for $y$ are also roots of the given polynomial.
I hope this helps. If you spot any typos here then please feel free to let me know and I'll correct them.
