On p.46 (or p. 43 in the 1st edition) of Huybrechts and Lehn book Geometry of Moduli Spaces of Sheaves, 2nd ed., they write:
Since $K$ is $A$-flat and $I \otimes_k F_0$ is annilated by $m_A$, there is a natural isomorphism $$\mathrm{Ext}^1_{X_A}(K, I \otimes_k F_0) \simeq \mathrm{Ext}^1_{X_s}(K_0, F_0) \otimes_k I.$$
I don't understand how they are getting this isomorphism and I was hoping someone can explain why (I bet it is fairly easy).
Here is roughly the setup in case you don't have the book handy:
Let $A'$ be an Artin local ring with residue field $k$ and maximal ideal $m_{A'}$, let $I$ be an ideal of $A'$ such that $m_{A'}I=0$, and let $A=A'/I$. Let $X=X_A$ be a projective variety over $\text{Spec } A$, and let $X_s=X_A \otimes k$ be the fiber over the closed point. Let $K,F$ be a $\mathcal{O}_X$-modules flat over $\text{Spec } A$, and let $K_0=K \otimes A/m$, $F_0=F \otimes A/m$ be the restriction to $X_s$. I think that explains all the notation.
Here is my attempt to understand the isomorphism: An element in $\mathrm{Ext}^1_{X_A}(K, I \otimes_k F_0)$ is given by a short exact sequence $$0 \to I \otimes_k F_0 \to B \to K \to 0$$
Since $K$ is $A$-flat, hence $\mathrm{Tor}^1_A(K, A/m)=0$ so apply $- \otimes_A A/m_A$ to get an exact sequence (exact on the left because $\mathrm{Tor}^1_A(K, A/m)=0$): $$0 \to I \otimes_k F_0 \to B_0 \to K_0 \to 0$$
This gives an element of $\mathrm{Ext}^1_{X_s}(K_0, I \otimes_k F_0)$. Here is why we can move the $I$ to the outside: Since $I$ is an ideal supported at the closed point of $A$ (since $m_AI=0$), it is a $k$-vector space, and so we can think of the $\mathcal{O}_{X_s}$-module $I \otimes_k F_0$ as $\dim_k I$ copies of $F_0$, and so we can pull the $I$ outside.
Conversely, given an extension $$0 \to I \otimes_k F_0 \to B_0 \to K_0 \to 0$$ I think I need to use the flatness of $K$ (or $F$) to get an extension $B$
$$0 \to I \otimes_k F_0 \to B \to K \to 0$$
but I don't know exactly how.
The exact sequence $$0\to I\otimes_kF_0\to B_0\to K_0\to 0$$ is also an exact sequence of ${\mathcal O}_{X_A}$ modules, and we can form its pullback with respect to the ${\mathcal O}_{X_A}$-linear surjection $K\to K_0$. This gives an exact sequence $$0\to I\otimes_kF_0\to B\to K\to 0$$ as desired, with $B$ being defined by the pullback property.