Consider any $x_2'$ and $x_2''$ in $\mathbb{R}_+$. Without loss of generality, let $x_2'' > x_2'$. We can choose $x_1'=f(x_2'')-f(x_2') > 0$ so that $U(0,x_2'')=U(x_1',x_2')$. Let $\lambda(x_1',x_2')+(1-\lambda)(0,x_2'')$ be a convex combination of $(x_1',x_2')$, and $(0,x_2'')$. Since $\succsim$ is strictly convex and $U(0,x_2'')=U(x_1',x_2')$,
\begin{eqnarray*}
&& U(\lambda(x_1',x_2')+(1-\lambda)(0,x_2'')) > U(x_1',x_2') \\
&\Rightarrow & U(\lambda x_1'+(1-\lambda)0,\lambda x_2'+(1-\lambda)x_2'') > \lambda U(x_1',x_2')+(1-\lambda)U(0,x_2'')\ldots(\because U(0,x_2'')=U(x_1',x_2')) \\
&\Rightarrow & \lambda x_1'+(1-\lambda)0 + f(\lambda x_2'+(1-\lambda)x_2'') > \lambda (x_1'+f(x_2'))+(1-\lambda)(0 + f(x_2''))\\
&\Rightarrow & \lambda x_1'+(1-\lambda)0 + f(\lambda x_2'+(1-\lambda)x_2'') > \lambda x_1'+(1-\lambda)0 + \lambda f(x_2')+(1-\lambda)f(x_2'') \\
&\Rightarrow & f(\lambda x_2'+(1-\lambda)x_2'') > \lambda f(x_2')+(1-\lambda)f(x_2'')\end{eqnarray*}
Therefore, $f(\cdot)$ is strict concave.