Replacing $y= 0$ in $f(x,t+y) = f(x,y) + tx $,
$f(x,t) = f(x,0) + tx$
i.e. $f(x,y) = f(x,0) + xy$ (equation $(1)$)
Replacing $x=0,y= 0$ in $f(x+t,y) = f(x,y) + ty $,
$f(t,0)=f(0,0) = K$
i.e. $f(x,0)=K$
putting $f(x,0)=K$ in equation $(1),$
$f(x,y) = K + xy$