= Solution
Let $u,v,w$ be the consecutive primitive <toric ray> vectors around the compact invariant curve $F=F_v$. Smoothness gives oriented determinants $\det(u,v)=\det(v,w)=1$, hence $u+w=bv$ for an integer $b$. Choose an <algebraic torus character> $m$ with $\langle m,u\rangle=0$ and $\langle m,v\rangle=1$. The <principal divisor on a toric variety> has coefficient one at $F$, coefficient $b$ at $F_w$, and zero at $F_u$. The other invariant curves do not meet $F$, while each adjacent one meets it transversely once. Intersecting this <principal divisor> with the proper curve $F$ gives $0=F^2+b$. This proves the <self-intersection formula for a toric surface divisor> even when the ambient surface is not complete.
If $F^2=-1$, then $b=1$ and $v=u+w$. The vectors $u,w$ form a lattice basis since $\det(u,w)=1$, and their <toric cone> is the union of the two <toric cones> adjacent to $v$. Deleting $v$ is consequently a smooth <toric fan> coarsening. Its inverse is precisely the <toric blowup at a torus-fixed point>, obtained by inserting the sum of the two primitive basis vectors. Conversely, that insertion gives $u+w=v$, and hence <algebraic self-intersection> $-1$. Thus the <toric contraction of a minus-one curve> satisfies
$$
\boxed{F\text{ contracts torically to a smooth point}\ \Longleftrightarrow\ F^2=-1.}
$$
For the cyclic quotient description, first take a strongly convex <toric cone> of dimension two. Write its primitive boundary vectors as $a,b$ and put $N_0=\mathbb Za+\mathbb Zb$, $r=[N:N_0]$. Since $a$ is primitive, extend it to a basis of $N$ and write $b=ua+rc$ with $\gcd(u,r)=1$. Thus $N/N_0$ is cyclic of order $r$. In the lattice $N_0$, the <toric cone> is smooth and its <toric variety> is $\mathbb A^2$. Passing to the larger lattice $N$ takes the finite quotient whose invariant <monomials> are precisely those in the smaller dual lattice $M=N^*\subset N_0^*$. Relative to the two boundary coordinates, $c=(b-ua)/r$, so a generator acts with weights $(-u,1)$ modulo $r$. Replacing that generator by a suitable power makes the first weight one and the second some $q$ with $0<q<r$ and $\gcd(q,r)=1$. Therefore, for the singular case,
$$
\boxed{U_\sigma\cong\mathbb A^2/\mu_r,\qquad\zeta\cdot(x,y)=(\zeta x,\zeta^q y).}
$$
The <cyclic quotient surface singularity> can equivalently be described by the first-quadrant <toric cone> in
$$
L=\mathbb Z^2+\mathbb Z\frac{(1,q)}r,\qquad L^*=\{(i,j)\in\mathbb Z^2:i+qj\equiv0\pmod r\}.
$$
Indeed its coordinate ring is $k[x^iy^j:i,j\ge0,\ i+qj\equiv0\pmod r]=k[x,y]^{\mu_r}$. Characteristic zero ensures the stated ordinary finite-group quotient interpretation.
The restriction to singular <toric cones> of dimension two is necessary for $r>1$. A smooth <toric cone> has $r=1$ and gives $\mathbb A^2$ with the trivial quotient. A <toric cone> of dimension one gives $\mathbb A^1\times k^*$, and the zero <toric cone> gives $(k^*)^2$. These latter surfaces cannot be a quotient of the displayed kind, since they have nonconstant invertible <regular functions>, whereas $k[x,y]^{\mu_r}$ has only constant units. Thus the blanket affine-surface wording requires these separate cases.
For the singular case, compute the <negative continued fraction> by the ceiling form of the <Euclidean algorithm>. Start with $a_0=r$, $a_1=q$, and, while $a_i>0$, set
$$
b_i=\left\lceil\frac{a_{i-1}}{a_i}\right\rceil,\qquad a_{i+1}=b_ia_i-a_{i-1}.
$$
Then $0\le a_{i+1}<a_i$, so the process terminates at $a_{s+1}=0$. The gcd is unchanged at each step, hence $a_s=1$. Since $a_{i-1}>a_i>0$, every $b_i\ge2$. Rearranging the recurrence gives
$$
\boxed{\frac rq=[b_1,\ldots,b_s]^-=b_1-\frac1{b_2-\dfrac1{\cdots-1/b_s}}.}
$$
Now define lattice vectors
$$
v_0=(0,1),\qquad v_1=\frac{(1,q)}r,\qquad v_{i+1}=b_iv_i-v_{i-1}.
$$
To verify this <Hirzebruch–Jung resolution> algorithm, write $v_i=(p_i,a_i)/r$ with $p_0=0$, $p_1=1$ and $p_{i+1}=b_ip_i-p_{i-1}$. Induction gives $a_i\equiv qp_i\pmod r$, so $v_i\in L$. It also gives $p_i a_{i+1}-a_i p_{i+1}=-r$. Thus every adjacent pair has absolute determinant $1/r$, the covolume of $L$, and is a basis of $L$. The $p_i$ increase strictly, while the $a_i$ decrease to zero, so these <toric rays> occur in order inside the first quadrant. At termination, $a_s=1$ and the determinant identity gives $p_{s+1}=r$, hence $v_{s+1}=(1,0)$. All boundary and inserted vectors are primitive because they occur in lattice bases.
Subdivide the quadrant by $v_1,\ldots,v_s$. Every resulting <toric cone> is smooth by the <smoothness criterion for a toric variety>, and the <fan subdivision> gives a proper birational <toric morphism>, hence a resolution. Each interior <toric ray> has a complete one-dimensional star, so its <exceptional curve of a surface resolution> is $\mathbb P^1$. The recurrence yields
$$
\boxed{F_i^2=-b_i\le-2,\qquad F_i\cdot F_{i+1}=1,\qquad F_i\cdot F_j=0\ (|i-j|>1).}
$$
This proves minimality: no <exceptional curve of a surface resolution> can be contracted to a smooth point by the <Castelnuovo contraction criterion>. There is also a direct <toric fan> check. For fixed $i$, the determinants of $v_i$ with $v_{i+h}$, measured in units of the lattice covolume, satisfy the same recurrence with coefficients at least two and initial values zero and one. They are at least $h$. Omitting any intervening <toric ray> therefore leaves a <toric cone> of determinant at least two, so no nontrivial coarsening is smooth. Thus this is the <minimal resolution of a cyclic quotient surface singularity>.
If all exceptional <algebraic self-intersections> are $-2$, all $b_i=2$ and induction gives $[2,\ldots,2]^-= (s+1)/s$. Since $r,q$ are coprime, this forces $r=s+1$, $q=s$. Conversely $r/(r-1)$ has exactly $r-1$ coefficients equal to two. Therefore
$$
\boxed{F_i^2=-2\text{ for every exceptional curve}\ \Longleftrightarrow\ q=r-1.}
$$
For the final <toric fan>, the minimal smooth <fan subdivision> of the <toric cone> from $(0,1)$ to $(r,1)$ inserts the <toric rays> $(j,1)$ for $1\le j<r$. Adjacent determinants have absolute value one, and $(j-1,1)+(j+1,1)=2(j,1)$, so its inserted curves have <algebraic self-intersection> $-2$ and it is the minimal <fan subdivision> just described. The second original <toric cone>, from $(r,1)$ to $(1,0)$, is already smooth.
Write $v_j=(j,1)$ and $e=(1,0)$. In the resolved <toric fan>, the rightmost interior <toric ray> obeys $v_{r-1}+e=v_r$, so its curve has <algebraic self-intersection> $-1$ and can be blown down to a smooth point. After deleting it, the same relation $v_{r-2}+e=v_{r-1}$ applies to the new rightmost interior <toric ray>. Continue deleting $v_r,v_{r-1},\ldots,v_1$ in that order. Each deletion is the inverse of a <toric blowup at a torus-fixed point>. The last <toric fan> has only the quadrant <toric cone> and defines $\mathbb A^2$. Consequently
$$
\boxed{X_{\Sigma'}\longrightarrow\mathbb A^2\text{ is the composite of exactly }r\text{ smooth-point toric blowdowns}.}
$$
All maps induce the identity on the dense <algebraic torus>, so this composite is the same morphism as $X_{\Sigma'}\to X_\Sigma\to\mathbb A^2$. The intermediate smooth <toric fans> need not include the singular <toric fan> $\Sigma$.
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