Let $R$ be a ring, commutative with $1$, subring of a ring $R'$. Let $\mathfrak{p}$ be an ideal of $R$. Let's denote by $\mathfrak{p}R'$ the extended ideal, i.e. the ideal generated by $\mathfrak{p}$ in $R'$.
EDIT: Assume that $R'$ is a free $R$-module of finite rank $n$ and let $x_1,\ldots,x_n$ be a basis of $R'$ over $R$. Consider the residue classes $\overline{x_1},\ldots,\overline{x_n}$ modulo $\mathfrak{p}R'$. I want to show that $\{\overline{x_i}\}$ is a basis for $R'/\mathfrak{p}R'$ over $R/\mathfrak{p}$.
This is a special case of the following general fact: if $R$ is a ring and $M$ is a free $R$-module with basis $\{m_i:i\in I\}$, then for any ring map $R\rightarrow S$, $M_S=S\otimes_RM$ is a free $S$-module with basis $\{1\otimes m_i:i\in I\}$. This amounts to the fact that tensor commutes with direct sums. In the situation asked about, $S=R/\mathfrak{p}$ and $M=R^\prime$, in which case $R/\mathfrak{p}\otimes_RR^\prime=R^\prime/\mathfrak{p}R^\prime$ as $R/\mathfrak{p}$-modules with $1\otimes x_i$ mapping to $x_i+\mathfrak{p}R^\prime$.
In the special case of interest this can be shown directly. It is clear that the $x_i+\mathfrak{p}R^\prime$ generate $R^\prime/\mathfrak{p}R^\prime$ over $R/\mathfrak{p}$. Suppose $\sum_i (r_i+\mathfrak{p})(x_i+\mathfrak{p}R^\prime)=0$, i.e., that $\sum_i r_i x_i\in\mathfrak{p}R^\prime$. Then $\sum_i r_i x_i=\sum_i s_ix_i$ with $s_i\in\mathfrak{p}$, so by $R$-linear independence, $r_i=s_i\in\mathfrak{p}$ for all $i$.
EDIT (to address the original, pre-edit question): The fact that every element of $\mathfrak{p}R^\prime$ has the form $\sum_i s_ix_i$ with $s_i\in\mathfrak{p}$ only uses that $R^\prime$ is generated as an $R$-module by the $x_i$. An element of $\mathfrak{p}R^\prime$ is of the form $\sum_j s_j r_j^\prime$ with $s_j\in\mathfrak{p}$ and $r_j^\prime\in R^\prime$ (this is a matter of definition). Now write $r_j^\prime=\sum_i r_{ij}x_i$ for $r_{ij}\in R$ (here we use that the $x_i$ generate $R^\prime$), and then multiply everything out, using that $s_jr_{ij}\in\mathfrak{p}$ for all $i,j$ because $\mathfrak{p}$ is an ideal of $R$.