$\color{Green}{Background:}$
$\textbf{Definition:}$ (Universal property of the kernel) Let $R$ be a commutative ring and $f:A\to B$ a morphism of $R-$modules. Recall that the $\textit{kernel}$ of $f$ is an $R-$module $\text{ker }(f)$ equipped with a morphism a morphism $i:\text{ker }(f)\to A$ ssatisfying the following $\textit{universal property}$ the composite $fi$ is the zero morphism and for any morphism $g::K\to A$ such that $fg=0,$ there exists a unique morphism $\bar{g}:K\to \text{ker}(f)$ making the triangle
$\begin{array} {c} K\\ \bar{g} \big\downarrow \searrow g \\ \text{ker }(f)\xrightarrow{i} A& \xrightarrow{f} & B\\ \end{array}\quad (1)$
commute.
$\textbf{Lemma:}$ Let $R$ be a commutative ring and
$$\begin{array}{ccccccccc} A & \xrightarrow{f} & B \\ \alpha\big\downarrow & & \big\downarrow\beta \\ A' & \xrightarrow{f'} & B' \\ \end{array}\quad (2)$$
a commutative square in $\textbf{Mod}_R.$
The restriction $f|_{\text{ker }(\alpha)}:\text{ker }(\alpha)\to \text{ker }(\beta)$ is the unique $R-$module homomorphism making the square
$$\begin{array}{ccccccccc} \text{ker }(\alpha) & \xrightarrow{f|_{\text{ker }(\alpha)}} & \text{ker }(\beta) \\ i_\alpha\big\downarrow & & \big\downarrow i_\beta \\ A & \xrightarrow{f} & B \\ \end{array}\quad (3)$$
commute, where $i_\alpha$ and $i_\beta$ denote the inclusions. Moreover, if $f$ is an injection, then so is $f|_{\text{ker }(\alpha)}.$
Proof. This follows easily from the universal property of the kernel. Consider the composite $\beta f i_\alpha.$ By the commutativeity of square in (2) and the fact that $i_\alpha$ is the inclusion of the kernel of $\alpha$ we have that
$beta f i_\alpha=f' \alpha i_\alpha=0$
Thus by the universal property of $\text{ker }(\beta),$ there exists a unique $R-$module homomorphism $\text{ker }(\alpha)\to \text{ker }(\beta) $ making the square
$$\begin{array}{ccccccccc} \text{ker }(\alpha) & \xrightarrow{} & \text{ker }(\beta) \\ i_\alpha\big\downarrow & & \big\downarrow i_\beta \\ A & \xrightarrow{f} & B \\ \end{array}\quad (4)$$
commute. The fact that the unique morphism $\text{ker }(\alpha)\to \text{ker }(\beta) $ is the restriction $f|_{\text{ker }(\alpha)}$ follows immediately from the definition of the restriction. The last thing to check is that if $f$ is an injection, then so is $f|_{\text{ker }(\alpha)}.$ To see this, notice that if $f$ is an injection, then since $i_\alpha$ is an inclusion, $fi_\alpha$ is also an injection, so the composite $i_\beta\circ f|_{\text{ker }(\alpha)}$ is an injection, which implies that $f|_{\text{ker }(\alpha)}$ is an injection.
$\color{Red}{Questions:}$
I am trying to understand the universal property for $\text{ker }(\beta)$ in the above proof. I understand if I match the information in the proof to the universal property of the kernel and the commutative triangle (1), would i have the following matches: $\text{ker }(f)=\text{ker }(\beta), K=\text{ker }(\alpha), \bar{g}=f|_{\text{ker }(\alpha)},$ $f$ remains as it is, and $i=i_\beta$. But what would the map $g$ be? Basically, I am trying to see how the universal property of the kernel is applied to make the proof of the lemma an easy task. I just have to match the right data from the data in the lemma to that of the definition.
Thank you in advance
The map $g$ in the universal property will be the composition $fi_\alpha$ in $(3)$ and the map $f$ will be $\beta$. Note $$\beta fi_\alpha = (\beta f)i_\alpha = (f'\alpha)i_\alpha = f'(\alpha i_\alpha) = f'0 = 0$$ so by the universal property you get a map from the domain of $fi_\alpha$ to $\mathrm{ker} \ \beta$, i.e., a map $\mathrm{ker} \ \alpha \to \mathrm{ker} \ \beta$.