$\DeclareMathOperator{\im}{Im}$
Let's do some grunt diagram chasing.

First, we prove $\overline{\delta}$ is injective. Suppose $x\in\ker\overline{\delta}$. Then $x\in\ker(\delta)=\im(\varepsilon)$, so $x=\varepsilon(k)$ for some $k\in K$. But also, $x\in\ker\varphi$, so
$$0=\varphi(x)=\varphi(\varepsilon(k))=\varepsilon'(\alpha(k))$$
But $\epsilon'$ is injective, so $\alpha(k)=0$. Since $\alpha$ is an isomorphism $k=0$, so $x=\varepsilon(k)=\varepsilon(0)=0$.

Therefore, $\overline{\delta}$ is injective.

Now let us prove $\overline{\delta}$ is surjective. Let $y\in\ker\psi$. Then $\psi(y)=0$, so
$$0=\eta'(\psi(y))=\beta(\eta(y))$$
Since $\beta$ is an isomorphism, $\eta(y)=0$, so $y\in\ker(\eta)=\im(\delta)$. Write $y=\delta(x)$ for some $x\in M$.

We have $\delta'(\varphi(x))=\psi(\delta(x))=\psi(y)=0$, so $\varphi(x)\in\ker\delta'=\im(\varepsilon')$. Let $k'\in K'$ such that $\varphi(x)=\varepsilon'(k')$. Since $\alpha$ is an isomorphism, there is $k\in K$ such that $k'=\alpha(k)$. Then
$$\varphi(\varepsilon(k))=\varepsilon'(\alpha(k))=\varepsilon'(k')=\varphi(x)$$
So now we use the $R$-module structure: Let $z=x-\varepsilon(k)$. Then the above means that $z\in\ker\varphi$. We prove that $\overline{\delta}(z)=y$:
$$\overline{\delta}(z)=\delta(x)-\delta\varepsilon(k)=\delta(x)=y$$
because $\delta\varepsilon=0$, since the first row is exact.

I can give some reasons why someone working with modules ought to care about these properties and abstract them into definitions.

For flat modules, the definition arises naturally when we intuit something from the notation, and then discover that this notation can sometimes be misleading.

For injective and projective modules, the definition comes from looking at how homomorphisms naturally arise from other homomorphisms in certain situations, and asking the question "Does every homomorphism arise in this natural way?"

**Flat**:

Let $P, M, N$ be modules over your commutative ring $R$. Suppose that $M$ is a submodule of $N$. The generators of $N \otimes_R P$ are written as $n \otimes p$ for $n \in N$ and $p \in P$ and likewise for the generators of $M \otimes_R P$.

There is a natural $R$-module homomorphism $\varphi: M \otimes_R P \rightarrow N \otimes_R P$ given on generators by

$$\varphi(m \otimes p) = m \otimes p$$

By the notation, it looks like $\varphi$ doesn't actually do anything. From the notation, and since $M$ is a subset of $N$, it looks like $M \otimes_R P$ should be a submodule of $N \otimes_R P$. But this isn't always true. You cannot guarantee this for all inclusions of submodules $M \subset N$ unless $P$ is flat. This is one possible definition you can take for flat modules: modules for which the above intuition always works.

**Injective**:

Easier to motivate this for abelian groups (that is, $\mathbb Z$-modules). Let $A$ be a subgroup of an abelian group $B$. Let $\mathbb C^{\ast}$ be the (multiplicative) group of complex numbers. If $\chi: A \rightarrow \mathbb C^{\ast}$ is a group homomorphism, one might wonder whether it is possible to extend $\chi$ (possibly nonuniquely) to a group homomorphism $\overline{\chi}: B \rightarrow \mathbb C^{\ast}$. It turns out that this is indeed possible, because $\mathbb C^{\ast}$ is an injective object in the category of abelian groups.

You can take this as the definition of injective: an abelian group $C$ is injective if whenever $A \subset B$ are abelian groups, every homomorphism of $A$ into $C$ can be extended (possibly nonuniquely) to a homomorphism of $B$ into $C$.

This is a pretty useful property for abelian groups to have. I ended up randomly needing the fact that $\mathbb C^{\ast}$ is an injective abelian group in my dissertation.

**Projective**:

Let $R$ be a commutative ring with identity, and let $I \subset J$ be ideals of $R$. There is a natural ring homomorphism

$$R/I \rightarrow R/J$$

$$r+I \mapsto r+J$$

Now let $B$ be another commutative ring with identity. If you're given a homomorphism of $B$ into $R/I$, you can compose with the natural homomorphism above and get a homomorphism of $B$ into $R/J$. It's natural to ask whether there are homomorphisms of $B$ into $R/J$ which don't arise in this fashion. This is the sort of idea that projective modules are based on.

Let $P$ be an $R$-module, and let $N$ be a submodule of an $R$-module $M$. You have a natural $R$-module homomorphism $\pi: M \rightarrow M/N$. If $\varphi: P \rightarrow M/N$ is an $R$-module homomorphism, you might ask the question of whether $\varphi$ always comes from a (possibly nonunique) $R$-module homomorphism of $P$ into $M$. The answer "Yes for all choices of $N \subset M$" is equivalent to $P$ being projective.

## Best Answer

I believe the statement is false: Let $A=B=C=B'=\mathbb{Z}\oplus \mathbb{Z}$ and $A'=C'=\mathbb{Z}$. Consider the following maps: $$ \psi(x,y)=(x,0), \quad \phi(x,y)=(0,y) $$ $$ \alpha(x,y)=x, \quad \beta(x,y)=(x,y), \quad \gamma(x,y)=y $$ $$ \psi'(x) = (x,0), \quad \phi'(x,y)=y. $$

Then it is easy to see that the diagram commutes, that $\alpha$ and $\gamma$ are surjective and $\beta$ is injective but $\gamma$ is not injective. Also, the rows are exact because $$ \operatorname{Img}(\psi) = \mathbb{Z}\oplus 0 = \ker(\phi) $$ and $$ \operatorname{Img}(\psi') = \mathbb{Z}\oplus 0 = \ker(\phi'). $$