Clearly the image of a torsion point is a torsion point, so it only remains to prove the surjectivity of  $f \colon A_{\mathrm{tors}} \to B_{\mathrm{tors}}$.

Up to a translation, we may assume that $f \colon A \to B$ is a group homomorphism.
Let $b \in B_{\mathrm{tors}}$ be a point of order $n$, namely $nb=0$. Then if $a \in A$ is any preimage of $b$ via $f$, we have $$f(na)=nf(a) =na=0,$$ hence $na \in \ker f$. 

But $\ker f$ is an abelian subvariety of $A$, in particular it is a divisible group; so we can find $a' \in \ker f$ such that $na = na'$, that is $n(a-a')=0.$ So $a-a' \in A_{\mathrm{tors}}$ and moreover $$f(a-a')=f(a)-f(a') = b-0= b,$$ 
i.e. $a-a'$ is a preimage of $b$ that lies in $A_{\mathrm{tors}}$. This shows the desired surjectivity.