I'll first label the maps as follows, for the rows we want
$$
0 \to A_i \xrightarrow{f_i} B_i \xrightarrow{g_i} C_i \to 0,
$$
and for the column I'll define
$$
0 \to A_1 \xrightarrow{f_A} A_2 \xrightarrow{g_A} A_3 \to 0,
$$
and similarly for the second and third columns
Now suppose that $c_1 \in C_1$, then there exists some $b_2 \in B_2$ such that $g_2(b_2) = f_C(c_1)$. Then
$$
g_3g_B(b_2) = g_Cg_2(b_2) = g_Cf_C(c_1) = 0,
$$
and so there exists some $a_3 \in A_3$ such that $f_3(a_3) = g_B(b_2)$. But then there exists $a_2 \in A_2$ such that $a_3 = g_A(a_2)$, and then
$$
g_Bf_2(a_2) = f_3g_A(a_2) = f_3(a_3) = g_B(b_2),
$$
and so $b_2 - f_2(a_2) \in \ker g_B = \operatorname{img} f_B$, and so there exists some $b_1 \in B_1$ such that $f_B(b_1) = b_2 - f_2(a_2)$. Then
$$
f_Cg_1(b_1) = g_2f_B(b_1) = g_2(b_2 - f_2(a_2)) = g_2(b_2) - g_2f_2(a_2) = g_2(b_2) = f_C(c_1),
$$
and so $g_1(b_1) = c_1$ since $f_C$ is injective. Hence $B_1 \xrightarrow{g_1} C_1$ is surjective.
EDIT: To add a comment on your final question. Suppose that we are given that $g_2f_2 = 0$ and that $f_C$ is monic. Then I claim that $g_1f_1 = 0$. Well it suffices to show that $f_Cg_1f_1 = 0$, but by the commutative diagram
$$f_Cg_1f_1 = g_2f_Bf_1 = g_2f_2f_A = 0,$$
and so we're done.
First of all, a convenient way to look at fibered products in this context is the following:
Denoting $\alpha_i: A_i \rightarrow A_0 , \;i=1,2$, the fibered product $A':=A_1\times_{A_0}A_2$ can be described as the kernel of the map $f_A:A_1 \oplus A_2 \stackrel{\begin{pmatrix}\alpha_1 \\ -\alpha_2\end{pmatrix}}{\longrightarrow} A_0$, so that one has an exact sequence
$$0 \rightarrow A' \rightarrow A_1 \oplus A_2 \rightarrow A_0,$$
and similarly for $B$'s and $C's$.
Now, your assumption implies that we have a commutative diagram with exact rows
\begin{array}
\\
0 & \rightarrow & A_1\oplus A_2 & \rightarrow & B_1\oplus B_2 & \rightarrow &C_1\oplus C_2
& \rightarrow & 0 \\
& & f_A \downarrow & & f_B \downarrow & & f_C \downarrow & & \\
0 & \rightarrow & \;\;\;\;A_0 & \rightarrow &\;\;\;\; B_0 & \rightarrow & \;\;\;\;C_0
& \rightarrow & 0, \\
\end{array}
So by the snake lemma, we obtain an exact sequence
$$0 \rightarrow A' \rightarrow B' \rightarrow C' \stackrel{\delta}{\rightarrow} \mathrm{Coker}\,f_A \cdots$$
So, the sufficient and necessary condition is that $\delta$ is the zero morphism. This is not easy to control for, so a perhaps-better condition, sufficient for exactness, is that $f_A$ is surjective. This is equivalent to $A_1 \rightarrow A_0$, $A_2 \rightarrow A_0$ being "jointly surjective" - meaning that the sum of images of the two maps is the whole group $A$.
(Note: This argument is actually not that far from justification for why the Mittag-Leffler conditions works for inverse ($\mathbb{N}$-indexed) limits.)
Best Answer
Yes. If the maps are displayed as $$ \require{AMScd} \begin{CD} 0 @>>> A_1 @>a_1>> A_2 @>{a_2}>> A_3 @>>> 0 \\ @. @V{f_1}VV @V{f_2}VV @V{f_3}VV \\ 0 @>>> B_1 @>>b_1> B_2 @>>{b_2}> B_3 @>>> 0 \end{CD} $$ try to prove that $f_i(\ker a_i) = \ker b_i$ and $f_{i+1}(\operatorname{im} a_i) = \operatorname{im} b_i$ for $i=1,2$.