= Solution
The overlap of the two charts is obtained by inverting $c=t^3$. Since $c^2=b^3$, this also inverts $b=t^2$; then $t=c/b$ and $t^{-1}=b/c$. Thus its coordinate <ring> is $R[t,t^{-1}]$. The other chart has coordinate $s=t^{-1}$. A <global section> is therefore an element in the intersection
$$
\Gamma(X,\mathcal O_X)=R[\pi t,t^2,t^3]\cap R[t^{-1}]=R
$$
inside $R[t,t^{-1}]$: a <Laurent polynomial> with both only nonnegative and only nonpositive exponents is constant.
Let $k=R/(\pi)$. Reducing the presentation of $B$ gives
$$
B_s=k[a,b,c]/(a^2,ab,ac,c^2-b^3).
$$
As a <vector space> and as a <ring> extension, this is
$$
B_s=k[b,c]/(c^2-b^3)\oplus ka,
$$
where $a^2=0$ and $a$ is annihilated by $b,c$. In particular $a\ne0$: the free <basis> of $B$ in part ii remains a <basis> after tensoring with $k$. Although $a$ mapped to $\pi t$ before reduction, reducing the subring is not the same operation as taking its image inside $k[t]$; that distinction is exactly where the extra section arises.
On the overlap, $c$ is invertible and $ac=0$, so $a$ restricts to zero. Therefore the section $a$ on the first chart glues to zero on the second chart and defines a nonzero global nilpotent section $\eta$ on $X_s$. More precisely the reduced first-chart <ring> is $k[t^2,t^3]$, and
$$
k[t^2,t^3]\cap k[t^{-1}]=k
$$
in $k[t,t^{-1}]$. The only extra compatible sections are multiples of $a$. Hence
$$
\boxed{\Gamma(X_s,\mathcal O_{X_s})\cong k[\eta]/(\eta^2),\qquad \Gamma(X,\mathcal O_X)=R.}
$$
The map printed in the question reduces $R$ modulo $\pi$ and includes its image as constants. Its kernel is $\pi R$ and its cokernel is $k\eta$. More strongly, even the usual base-change map after tensoring,
$$
\boxed{\Gamma(X,\mathcal O_X)\otimes_Rk=k\longrightarrow k[\eta]/(\eta^2),}
$$
is not surjective. This is <global-section base-change failure in a flat projective family>, despite the properness and flatness proved above.
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