First of all, notice that writing $a_{(1)} = a_{(1)}'$ does not properly make sense. The Heyneman-Sweedler sigma notation $\Delta(a) = \sum a_{(1)} \otimes a_{(2)}$ is a easy-to-handle way of denoting the more appropriate $\Delta(a) = \sum_{i=1}^{n_a} b_i \otimes c_i$.
Despite of this, there is an easier way to show that 1 implies 2, which is that of exhibiting an explicit inverse for $T_1$ and $T_2$. I would suggest you to try with
$$T_1': B \otimes B \to B \otimes B, \qquad a \otimes b \mapsto \sum a_{(1)} \otimes S\left(a_{(2)}\right)b$$
for example, where $S : B \to B$ is the antipode.
Once you proved that 1 implies 2, proving that 2 implies 1 becomes much easier. Call, for example, $T_1'$ the inverse of $T_1$. Since $T_1$ is right $B$-linear with respect to the regular right $B$-action on both the domain and the codomain, $T_1'$ has to satisfy the same property and hence
$$T_1'(a \otimes bc) = T_1'(a \otimes b)\cdot (1 \otimes c).$$
By looking at the inverse of $T_1$ from the previous implication, it comes natural to define
$$S := (\varepsilon \otimes B) \circ T_1' \circ (B \otimes u): B \to B$$
where $u : \Bbbk \to B$ is the unit of $B$ as a $\Bbbk$-algebra. It follows that
$$
\begin{aligned}
\sum S(a_{(1)})a_{(2)} & = \sum (\varepsilon \otimes B)\big(T_1'(a_{(1)}\otimes 1)\big)a_{(2)} \\
& = \sum (\varepsilon \otimes B)\big(T_1'(a_{(1)}\otimes 1)\cdot(1 \otimes a_{(2)})\big) \\
& = (\varepsilon \otimes B)\left(T_1'\left(\sum a_{(1)}\otimes a_{(2)}\right)\right) \\
& = (\varepsilon \otimes B)\left(T_1'\left(T_1(a \otimes 1)\right)\right) = \varepsilon(a)1,
\end{aligned}
$$
which is one of the two identities required to have an antipode. I am sure that now you may imagine how to use the inverse of $T_2$ to complete the proof.
Therefore, let me conclude by performing a Hopf-algebraic virtuosity.
In fact, the invertibility of $T_1$ alone already suffices to prove that $S$ defined above is an antipode. The reason is that $T_1$ is also colinear with respect to the regular left coactions on both domain and codomain, whence
$$(\Delta \otimes B)\big(T_1'(a \otimes b)\big) = \sum a_{(1)} \otimes T_1'(a_{(2)} \otimes b).$$
If now we apply $B \otimes \varepsilon \otimes B$ to both sides, we find that
$$T_1'(a \otimes b) = \sum a_{(1)} \otimes (\varepsilon \otimes B) \left(T_1'(a_{(2)} \otimes b)\right)$$
and so
$$a \otimes 1 = T_1T_1'(a \otimes 1) = \sum a_{(1)} \otimes a_{(2)}(\varepsilon \otimes B) \left(T_1'(a_{(3)} \otimes 1)\right) = \sum a_{(1)} \otimes a_{(2)}S\left(a_{(3)}\right).$$
A further application of $\varepsilon \otimes B$ entails that
$$\varepsilon(a)1 = \sum a_{(1)}S\left(a_{(2)}\right).$$
