The first step is correct, but the second step is not: You cannot say anything about the kernel after tensoring. Also note that your second step is purely formal and would apply to every additive functor which preserves epis. But not every such functor is right exact.
Let $M_1 \to M_2 \to M_3 \to 0$ be an exact sequence. We want to show that, for every module $N$, the sequence $M_1 \otimes N \to M_2 \otimes N \to M_3 \otimes N \to 0$ is exact, i.e. that $M_2 \otimes N \to M_3 \otimes N$ is a cokernel of $M_1 \otimes N \to M_2 \otimes N$. This means, by the universal property of the cokernel, that for every "test" module $T$, the sequence $0 \to \hom(M_3 \otimes N,T) \to \hom(M_2 \otimes N,T) \to \hom(M_1 \otimes N,T)$ is exact (as abelian groups, but then also as modules). By definition of the tensor product, this sequence is isomorphic to the sequence $0 \to \mathrm{Bilin}(M_3,N;T) \to \mathrm{Bilin}(M_2,N;T) \to \mathrm{Bilin}(M_1,N;T)$. $(\star)$
Thus, the claim is actually equivalent to a statement about bilinear maps. And this can be checked now directly. I will leave out the trivial steps. For the only interesting one, let $\beta : M_2 \times N \to T$ be a bilinear map which vanishes on $M_1 \times N$. Define $\gamma : M_3 \times N \to T$ as follows: If $m_3 \in M_3$, $n \in N$, choose a preimage $m_2 \in M_2$ of $m_3$ and define $\gamma(m_3,n):=\beta(m_2,n)$. This is well-defined, because every other choice of $m_2$ is of the form $m_2+x$ for some $x$ coming from $M_1$, and then $\beta(m_2+x,n)=\beta(m_2,n)+\beta(x,n)=\beta(m_2,n)$. One sees directly that $\gamma$ is bilinear because $\beta$ is. And course $\gamma$ is the desired preimage in $\mathrm{Bilin}(M_3,N;T)$.
This is not the most conceptual proof. You have already mentioned the one using adjoint functors. But we can also choose an alternative ending for the proof above: The sequence $(\star)$ is isomorphic to $0 \to \hom(N,\hom(M_3,T)) \to \hom(N,\hom(M_2,T)) \to \hom(N,\hom(M_1,T))$, which is exact because $\hom(N,-)$ is left exact and $\hom(-,T)$ is right exact.
And yet another ending (which explains Qiaochu's comment): The isomorphism $\mathrm{Bilin}(-,N;T) \cong \hom(-,\hom(N,T))$ shows that this functor is representable and therefore right exact, hence $(\star)$ is exact.
The proof as intended by Weibel doesn't seem to work, since it requires one to solve the related question asked here Are those two ways to relate Extensions to Ext equivalent?. However, by using the dual version of his proof we can avoid this issue:
Pick an exact sequence $0\to B \to I\xrightarrow{\pi} N\to 0$, where now $I$ is an injective object. Then apply $Ext(A,-)$ to obtain en exact sequence
$$ ... \to Hom(A,N) \xrightarrow{\partial} Ext(A,B) \to 0,$$ and pick $\gamma \in Hom(A,N)$ with $\partial(\gamma)=x$. Now we let $X$ be the pullback of $A\xrightarrow{\gamma} N \xleftarrow{\pi}I$. This fits into a commutative diagram with exact rows:
\begin{array}{ccccccccc}
0 & \xrightarrow{} & B & \xrightarrow{} & X & \xrightarrow{} & A & \xrightarrow{} & 0\\
& & \parallel & & \downarrow & & \downarrow & & \\
0 & \xrightarrow{} & B & \xrightarrow{} & I & \xrightarrow{} & N & \xrightarrow{} & 0.
\end{array}
The upper row is now an extension $\xi$ for which one directly sees that $\Theta(\xi)=x$:
Applying $Ext(A,-)$ again gives a long ladder diagram, from which we consider the square
\begin{array}{ccc}
Hom(A,A) & \xrightarrow{\partial'} & Ext^1(A,B) \\
\downarrow& & \parallel \\
Hom(A,N) &\xrightarrow{\partial} & Ext^1(A,B)
\end{array}
The $\partial$ here is the same as above, and by the definition in Weibel we have $\Theta(\xi) =\partial'(id_A)$. Finally, the left vertical arrow is composition with $\gamma$ by definition of the $Hom$ functor. So $$\Theta(\xi) =\partial'(id_A)=\partial(\gamma\circ id_A) =x.$$
Best Answer
At a minimum, there are only two things left to prove to complete the equivalences.
I'll prove (1) $\implies$ (3) and (4) $\implies$ (2), since you know (3) $\implies$ (4) and (2) $\implies$ (1).
I'm going to state two facts about $\newcommand\Ext{\operatorname{Ext}}\newcommand\Hom{\operatorname{Hom}}\Ext$, and if you're not familiar with them, then I'd suggest looking into these, because they're a bit beyond the scope of an answer to reprove here.
Fact 1 gives (1) $\implies$ (3), since $B\to 0 \to 0 \to 0 \to \cdots $ is already an injective resolution of $B$ when $B$ is injective, so $$\Ext^n(A,B) \cong H^n(\Hom(A,B)\to 0 \to 0 \to 0 \to 0),$$ so $\Ext^i(A,B)=0$ for $i>0$ (and any $A$).
Fact 2 gives (4) $\implies$ (2), since if $\Ext^1(A,B)=0$ for all $A$, then for any short exact sequence $0\to A'\to A\to A''\to 0$, we have the long exact sequence $$0\to \Hom(A'',B)\to \Hom(A,B)\to \Hom(A',B)\to \Ext^1(A'',B)=0,$$ since we assumed that $\Ext^1(A'',B)=0$ for any $A''$, so $\Hom(-,B)$ is an exact functor.
Edit:
Also I missed this when reading your question, but I realized I didn't directly address your first question about proving exactness in the middle when proving (1) $\implies$ (2).
This also follows from Fact 2 above, but that's overkill, there is actually an elementary proof that for any short exact sequence $\newcommand\toby\xrightarrow 0\to A'\toby{f} A\toby{g} A''\to 0$, and any $B$, the sequence $$0\to \Hom(A'',B)\toby{g^*} \Hom(A,B)\toby{f^*} \Hom(A',B)$$ is exact. Then $B$ being injective is equivalent to the last map being surjective for all short exact sequences.
Proof
Since $gf=0$, we have $f^*g^*=(gf)^*=0$, which means $\newcommand\im{\operatorname{im}}\im g^*\subseteq \ker f^*$ so we have two things to prove: (a) injectivity of $g^*$, and (b) that $\ker f^*\subseteq \im g^*$.
(a) If $\phi : A''\to B$ is some map, and $g^*\phi = \phi\circ g =0$, then, if $x\in A''$ is any element, since $g$ is surjective, $x=g(a)$ for some $a\in A$, so $\phi(x) = \phi(g(a))=0$. Therefore $\phi=0$, so $g^*$ is injective.
(b) Suppose $\phi : A\to B$ is in the kernel of $f^*$, so $\phi\circ f =0$. Then since $f$ is injective and $g$ is surjective, we can regard $A'$ as a submodule of $A$ and we have that $A''\cong A/A'$. Then $\phi : A\to B$ is a morphism that is zero on $A'$, so we know that it induces a morphism $\phi' : A''\to B$ defined by $\phi'(g(a))=\phi(a)$. But this is exactly what it means to say $\phi = g^*\phi'$, so $\phi$ is in the image of $g^*$. $\blacksquare$