No. For example, let $X = T = C$ be a smooth projective curve (but we keep the notation $X \times T$ to indicate the separate functions of the two). Let $E = \mathcal O_{\Delta_C}$, and let $E' = \mathcal O_{x \times T}$ for a point $x \in X$.

Since both the maps $\Delta_C \to X \times T \to T$ and $x \times T \to X \times T \to T$ are isomorphisms, we see that $E$ and $E'$ are flat over $T$. They also admit a presentation of the form you describe: we can take $A = A' = \mathcal O_{X \times T}$, with $K = \mathcal O_{X \times T}(-\Delta_C)$ and $K' = \mathcal O_{X \times T}(-x \times T)$. The sequences
$$\begin{array}{ccccccccc}0 & \to & \mathcal O_{X \times T}(-\Delta_C) & \to & \mathcal O_{X \times T} & \to & \mathcal O_{\Delta_C} & \to & 0\\0 & \to & \mathcal O_{X \times T}(-x\times T) & \to & \mathcal O_{X \times T} & \to & \mathcal O_{x \times T} & \to & 0\\\end{array}$$
are both exact, hence remain so in each fibre $X \times t$ (e.g. use right exactness of $-\otimes \mathcal O_{X \times t}$ plus the fact that the left hand side is locally free of rank $1$, so that the first map is injective by a rank count).

But $E|_{X \times t} = \mathcal O_t$ and $E'|_{X \times t} = \mathcal O_x$ (with the identification $X \times t \cong X$). Hence, $\mathcal Hom(E,E')$ jumps at $x = t$: there are no homomorphisms when $t \neq x$, but there is one when $t = x$.