Choose a presentation $A = R[x_1, \ldots, x_n]/I$ and write $I = (f_1, \ldots, f_m)$. Define the $A$-module $K$ by the short exact sequence $$ 0 \to K \to A^{\oplus m} \to I/I^2 \to 0 $$ where the $j$th basis vector $e_j$ in the middle is mapped to the class of $f_j$ on the right. Set $$ C = \text{Sym}^*_A(I/I^2). $$ The retraction is just the projection onto the degree $0$ part of $C$. We have a surjection $R[x_1, \ldots, x_n, y_1, \ldots, y_m] \to C$ which maps $y_j$ to the class of $f_j$ in $I/I^2$. The kernel $J$ of this map is generated by the elements $f_1, \ldots, f_m$ and by elements $\sum h_j y_j$ with $h_j \in R[x_1, \ldots, x_n]$ such that $\sum h_j e_j$ defines an element of $K$. By Algebra, Lemma \ref{algebra-lemma-exact-sequence-NL} applied to $R \to A \to C$ and the presentations above and More on Algebra, Lemma \ref{more-algebra-lemma-cotangent-complex-symmetric-algebra} there is a short exact sequence \begin{equation} \label{equation-sequence} I/I^2 \otimes_A C \to J/J^2 \to K \otimes_A C \to 0 \end{equation} of $C$-modules. Let $h \in R[x_1, \ldots, x_n]$ be an element with image $a \in A$. We will use as presentations for the localized rings $$ A_a = R[x_0, x_1, \ldots, x_n]/I' \quad\text{and}\quad C_a = R[x_0, x_1, \ldots, x_n, y_1, \ldots, y_m]/J' $$ where $I' = (hx_0 - 1, I)$ and $J' = (hx_0 - 1, J)$. Hence $I'/(I')^2 = A_a \oplus (I/I^2)_a$ as $A_a$-modules and $J'/(J')^2 = C_a \oplus (J/J^2)_a$ as $C_a$-modules. Thus we obtain \begin{equation} \label{equation-sequence-localized} C_a \oplus I/I^2 \otimes_A C_a \to C_a \oplus (J/J^2)_a \to K \otimes_A C_a \to 0 \end{equation} as the sequence of Algebra, Lemma \ref{algebra-lemma-exact-sequence-NL} corresponding to $R \to A_a \to C_a$ and the presentations above. \medskip\noindent Next, assume that $a \in A$ is such that $A_a$ is a local complete intersection over $R$. Then $(I/I^2)_a$ is finite projective over $A_a$, see More on Algebra, Lemma \ref{more-algebra-lemma-quasi-regular-ideal-finite-projective}. Hence we see $K_a \oplus (I/I^2)_a \cong A_a^{\oplus m}$ is free. In particular $K_a$ is finite projective too. By More on Algebra, Lemma \ref{more-algebra-lemma-transitive-lci-at-end} the sequence (\ref{equation-sequence-localized}) is exact on the left. Hence $$ J'/(J')^2 \cong C_a \oplus I/I^2 \otimes_A C_a \oplus K \otimes_A C_a \cong C_a^{\oplus m + 1} $$ This proves (1). Finally, suppose that in addition $A_a$ is smooth over $R$. Then the same presentation shows that $\Omega_{C_a/R}$ is the cokernel of the map $$ J'/(J')^2 \longrightarrow \bigoplus\nolimits_i C_a\text{d}x_i \oplus \bigoplus\nolimits_j C_a\text{d}y_j $$ The summand $C_a$ of $J'/(J')^2$ in the decomposition above corresponds to $hx_0 - 1$ and hence maps isomorphically to the summand $C_a\text{d}x_0$. The summand $I/I^2 \otimes_A C_a$ of $J'/(J')^2$ maps injectively to $\bigoplus_{i = 1, \ldots, n} C_a\text{d}x_i$ with quotient $\Omega_{A_a/R} \otimes_{A_a} C_a$. The summand $K \otimes_A C_a$ maps injectively to $\bigoplus_{j \geq 1} C_a\text{d}y_j$ with quotient isomorphic to $I/I^2 \otimes_A C_a$. Thus the cokernel of the last displayed map is the module $I/I^2 \otimes_A C_a \oplus \Omega_{A_a/R} \otimes_{A_a} C_a$. Since $(I/I^2)_a \oplus \Omega_{A_a/R}$ is free (from the definition of smooth ring maps) we see that (2) holds.