If $f^2(t) \le 1+2\int_0^tf(s)\mathrm{d}s$ prove $f(t)\le 1+t$

You are almost done. You correctly derived that $$ \sqrt{1+2\int_0^tf(s)\mathrm{d}s} - 1 \le t \, . $$ Now use the initial given inequality to conclude that $$ f^2(t) \le 1+2\int_0^tf(s)\mathrm{d}s \le (1+t)^2 \implies f(t) \le 1+t \, . $$