Following C. C. Rousseau and J. Sheehan "A class of ramsey problems involving trees" Journal of London Math. Soc. $1978$, I have some dudes about their proof for $R(K_l + \overline{K_t}, T_s) \le l(s-1)+t$.
Their proof uses induction on $s$. Suppose we use blue/red colorations of $K_p$ (with $p=l(s-1)+t$). Case $s=2$ holds if and only if any 2-coloration of $K_p$ have at least a red edge, but What if I color all edges of $K_p$ blue ?.
Suppose $(F_1,F_2)$ is a blue/red factorization of $K_p$ such that $K_l + \overline{K_t}$ is not a subgraph of $F_1$ neither $T_s$ is a subgraph of $F_2$, consider $T_{s-1} = T_s - v$ for some end vertex $v$ of $T_s$. By induction hypothesis $T_{s-1}$ is a subgraph of $F_2$, then exist a vertex $x$ in $V(T_{s-1})$ such that $xy \in E(F_1)$ for every $y \in V(K_p) - V(T_{s-1})$ (note $\vert V(K_p) - V(T_{s-1})\vert = (l-1)(s-1)+t$).
Then, if $l=1$ necessarily $x$ is adjacent to the $t$ vertices outside $T_{s-1}$ with edges in $F_1$, but What if these $t$ vertices have $F_1$ edges among them? I think this could be possible but then this not produces $K_{1,t}$ as a subgraph of $F_1$.
Thanks in advance