The first one is what you get if you integrate the equation that describes constant acceleration twice. If the acceleration is constant (=a), we have
$$x''(t)=a$$ This implies that there's a constant ##v_0## such that $$x'(t)=at+v_0.$$ The constant is denoted by ##v_0## because ##x'(t)## is the velocity at time t, and the result we just found implies that ##x'(0)=v_0##, so ##v_0## is the velocity at time 0. The result we found also implies that there's a constant ##x_0## such that
$$x(t)=\frac 1 2 at^2+v_0t+x_0.$$ The constant is denoted by ##x_0## because it's equal to ##x(0)##, the position at time 0.
The other one follows from the result for the velocity above, and
$$v(t)-v_0=\int_0^t x''(t)dt=at.$$ These results imply that
$$v(t)^2=(v_0+at)^2=v_0{}^2+a^2t^2+2v_0at=v_0{}^2+2a\bigg(\frac 1 2 at^2+v_0t\bigg)=(v_0)^2+2a(x(t)-x_0).$$
I think the formula you wrote down only holds if ##v(t_i)=v_0##, so you may want to check your book to see if it says exactly what you wrote, or something slightly different.