Roland left an excellent comment reminding me that the addition in an Abelian category (well technically category with finite biproducts) can be defined purely in terms of the direct sum, diagonal and codiagonal maps.
If $f,g : A\to B$ are morphisms, we define $f+g$ to be the composite
$$ A\newcommand\toby\xrightarrow\toby{\Delta_A} A \oplus A\toby{f\oplus g}B\oplus B \toby{\nabla_B} B,$$
where $\Delta_A : A\to A\oplus A$ is the diagonal map and $\nabla_B : B\oplus B\to B$ is the codiagonal.
Then clearly any functor which satisfies $T(f\oplus g)=Tf\oplus Tg$ and $T\Delta_A = \Delta_{TA}$ and $T\nabla_B = \nabla_{TB}$ is additive.
Then as long as $T$ preserves the biproduct (meaning $T(A\oplus B)=TA\oplus TB$ and $T\iota_X=\iota_{TX}$ and $T\pi_X = \pi_{TX}$ for the structure morphisms $\iota_A,\iota_B,\pi_A,$ and $\pi_B$), and $T$ preserves $0$, meaning that $T0 = 0$, $T$ has these properties. (Equality here means up to natural isomorphism.)
This is because $\Delta_A$ is the map such that $\pi_{A,i}\Delta_A=\newcommand\id{\operatorname{id}}\id_A$ for $i=1,2$,
and then taking $T$ of this, we have that $T\Delta_A$ satisfies $\pi_{TA,i}T(\Delta_A)=\id_{TA}$ for $i=1,2$, thus $T\Delta_A=\Delta_{TA}$. A dual argument shows that $T\nabla_B=\nabla_{TB}$, and finally $f\oplus g$ is defined to be the map such that
$$\pi_{B,1}(f\oplus g)\iota_{A,1} = f,$$
$$\pi_{B,2}(f\oplus g)\iota_{A,2} = g,$$
$$\pi_{B,1}(f\oplus g)\iota_{A,2} = 0,$$
and
$$\pi_{B,2}(f\oplus g)\iota_{A,1} = 0.$$
Again, it's not hard to see that the preservation of 0 and the biproduct ensures that $T(f\oplus g)$ satisfies the requirement to be $Tf\oplus Tg$.
Finally, if $T$ is left exact, then $T$ preserves $0$ and the biproduct, and thus is additive
(As indicated in my question, I've already proved this, but I figured I'd add it for any future readers)
$T$ preserves $0$
Depending on how you interpret the definition, this is either definitional, (since you can read the definition as implying that you take $T$ of the entire left exact sequence to get the resulting exact sequence) or must be proved, if you take it to mean only that extending the sequence $TA\to TB \to TC$ by zero on the left results in an exact sequence, but it turns out that there's no difference.
Consider
$$ 0 \to 0 \toby{\id_0} 0 \toby{\id_0} 0.$$
This is clearly exact,
so
$$ 0 \to T0 \toby{\id_{T0}} T0 \toby{\id_{T0}} T0$$
is exact, which implies that $$0=\ker \id_{T0} = T0=\operatorname{im}_{\id_{T0}},$$
as desired.
$T$ preserves the biproduct
Proof:
We know $\pi_A\iota_A =\id_A$, so $(T\pi_A)(T\iota_A)=\id_{TA}$. This also applies to $\pi_B\iota_B$, so $T\pi_B$ must be an epimorphism. Hence, exactness of
$$0\to A \toby{\iota_A} A\oplus B \toby{\pi_B} B \to 0$$
not only implies exactness of
$$0\to TA \toby{T\iota_A} T(A\oplus B) \toby{T\pi_B} TB, $$
but actually the exactness of
$$0\to TA \toby{T\iota_A} T(A\oplus B) \toby{T\pi_B} TB \to 0,$$
and the fact that $(T\pi_A)(T\iota_A)=\id_{TA}$ implies that this exact sequence is split, giving
$T(A\oplus B) \simeq TA\oplus TB$.
Best Answer
You have the right idea, but you really need the bottom row to be exact at $K$ and not necessarily at $A$.
Indeed, to get the factorisation of $U\to A_j$ through $K_j$ you need to prove that the composition $U\to A_j\to B_j$ is zero, and for this you only need to know that $K\to A\to B$ is zero, which is weaker than exactness at $A$.
On the other hand, to prove that the second leg of the pullback commutes, you use the uniqueness of the factorisation through $K$, or in other words, the fact that $K\to A$ is a monomorphism, which is equivalent to the exactness of $0\to K\to A$ at $K$.