Here we take the infinity category of simplicial ring $SCRing=Fun^{\prod}(Poly^{op},Spc)$ and follow the construction 25.3.1.1 in DAG by Lurie, where we extend the construction of square zero extension from $(A,M)\rightarrow A\oplus M$ with $A$ a polynomial ring and $M$ finite free $A$ module to $SCRMod^{cn}$.
Recall in the discrete case, for a ring morphism sequence $A\rightarrow C\stackrel{f}{\longrightarrow} B$, and a $B$ module $M$, we denote the pullback $f^{*}M$ as $M^{\sim}$ which action is defined by $c\cdot m=f(c)\cdot m$. Then we have a bijection: \begin{equation} \mathbf{Hom_{Alg_{A//B}}(C,B\oplus M)\simeq Hom_{Alg_{A//C}}(C,C\oplus M^{\sim})} \end{equation}
Indeed, consider a $A//C$ algebra morphism $C\rightarrow C\oplus M^{\sim}$, it implicitly requires that this is a section of the projection of the square zero extension in that the composition $C\rightarrow C\oplus M^{\sim}\stackrel{projection}{\longrightarrow} C$ is the identity map on $C$, thus $c\in C\rightarrow (\psi(c),d(c))\in C\oplus M^{\sim}$ satisfies $\psi=id$ and $d$ satisfies $c\cdot d(c')+c'\cdot d(c)=d(c\cdot c')$ for any $c,c'\in C$, this relation can naturally be transferred to a $A//B$ algebra morphism $C\rightarrow B\oplus M$ which is of form $c\in C\rightarrow (f(c),d(c))\in B\oplus M$. The bijection between these tow Hom-sets is easily verified by definition.
While in the derived setting, things become subtle because we need to check higher homotopy information instead of just points, moreover, by the construction of square zero extension by Lurie, we indeed can think of these simplical rings and their modules as some sifted colimits of discrete rings and modules, but even though we still can't transfer the bijection of Hom-sets into homotopy equivalences of mapping spaces:
\begin{equation} \mathbf{Map_{Alg_{A//B}}(C,B\oplus M)\simeq Map_{Alg_{A//C}}(C,C\oplus M^{\sim})} \end{equation}
Indeed, colimits don't commute with the mapping space in the second variable, and even if we can show pointwise bijection of the space morphism $\text{Map}_{Alg_{A//C}}(C,C\oplus M)\rightarrow \text{Map}_{Alg_{A//B}}(C,B\oplus M)$, this doesn't guarantee the homotopy equivalence of topological spaces. But I do believe the above equivalence is correct. Is there any categorical way to show this equivalence?
$\textbf{Motivation}$: This question is related with this post about the infinite suspension of $A\rightarrow B\in Alg_{/B}$. As suggested by comments in Achim Krause's answer, 'infinite suspension of pointed objects $A\rightarrow B\stackrel{f}{\longrightarrow} A$ is exactly the cotangent fiber $f_{*}L_{B/A}$.' Indeed, by Remark 7.4.3.16 in HA, $M\in Mod(A)=Sp(Alg^{aug}_{A})\rightarrow A\oplus M$ is exactly the infinite loop, thus it suffices to prove the adjunction $\text{Map}_{Mod(A)}(f_{*}L_{B/A}, M)\cong \text{Map}_{Alg_{A//A}}(B,A\oplus M)$. As $f_{*}$ is the left adjoint of the pullback $f^{*}$, we have $\text{Map}_{Mod(A)}(f_{*}L_{B/A},M)\cong \text{Map}_{Mod(B)}(L_{B/A},f^{*}{M}=M^{\sim})\cong \text{Map}_{Alg_{A//B}}(B,B\oplus M^{\sim})$ where the last equivalence is due to the universal property of the relative cotangent complex $L_{B/A}$. Thus it suffices to show $\text{Map}_{Alg_{A//A}}(B,A\oplus M)\cong \text{Map}_{Alg_{A//B}}(B,B\oplus M^{\sim})$, which is exactly the equivalence I want to prove in this post.
