The following question is taken from Arrows, Structures and Functors the categorical imperative by Arbib and Manes, "Advanced Modern Algebra" 2nd edition by Rotman, and "Representation theory of the Virasoro Algebra" by Iohara and Koga.
$\color{green}{Background}$
From Iohara and Koga's Representation theory of the Virasoro Algebra.
$\textbf{(1):}$ A pair $(K,i)$ is called the $\textit{kernel}$ of $f$ if it satisfies
i. $f\circ i=0$ and
ii. for any morphism such that $f\circ j=0$ there exists a unique morphism $k$ such that $j=i\circ k.$
In such case, the object $K$ is often denoted by $\text{ker }f$
From Rotman's Advanced Abstract Algebra 2nd edition.
From Rotman's Advanced Abstract Algebra 2nd edition.
$\textbf{(2):}$ Definition: Given two morphisms $f:B\rightarrow A$ and $g:C\rightarrow A$ in a category $\textbf{C},$ a $\textbf{solution}$ is an ordered triple $(D,\alpha,\beta)$ making the left-hand diagram commute. A $\textbf{pullback}$ (or $\textit{fibered product}$ is a solution $(D,\alpha,\beta)$ that is "best" in the following sense: for every solution $(X,\alpha',\beta'),$ there exists a unique morphism $\theta:X\rightarrow D$ making the right-hand diagram below commute.
Showing that kernel is a pullback in the case of $_R\textbf{Mod}.$
Suppose $f:B\rightarrow A$ is a homomorphism in $_R\textbf{Mod},$ then the pullback of the first diagram below is $({ker}f,0,i),$ where $i:{ker}f\rightarrow B$ is the inclusion. Let $i':X\rightarrow B$ be a map with $f\circ i'=0;$ then $(f\circ i')(x)=0$ for all $x\in X,$ and so $i'(x)\in {ker}f.$ If we define $\theta:X\rightarrow {ker}f$ to be the map obtained from $i'$ by changing its target, then the diagram commutes: $i\circ \theta =i'.$ To prove uniqueness of the map $\theta,$ suppose that $\theta':X\rightarrow {ker}f$ satisfies $i\circ\theta'=i'.$ Since $i$ is the inclusion, $\theta'(x)=i'(x)=\theta(x)$ for all $x\in X,$ and so $\theta'=\theta.$ Thus, $({ker}f,0,i)$ is a pullback.
From Arbib and Manes "Categorical Imperative"
$\textbf{(3) Proposition:}$ In the category $\textbf{Vect}$
$\textbf{(i)}$ $f$ is an epimorphism $\quad\text{iff}\quad$ $f$ is onto $\quad\text{iff}\quad$ $f$ is a coequalizer
$\textbf{(ii)}$ $f$ is a monomorphism $\quad\text{iff}\quad$ $f$ is one-to-one $\quad\text{iff}\quad$ $f$ is an equalizer
$\textbf{(4):}$ Definition: Let $\textbf{K}$ be a category with zero object $0.$ Then the $\textbf{kernel}$ $u:K\to A$ of $f:A\to B$ is given by the pullback diagram
$\color{Blue}{Exercise:}$ Let $f:X\to Y$ be a linear map in $\textbf{Vect}.$ Prove that the kernel of $f$ as in $(2)$ is the usual construction of the subspace $\{x\in X\mid f(x)=0\}.$
$\color{Red}{Solution:}$
Proof: We relabel the diagram in $\textbf{(2)}$ as follow:
$\begin{array}{ccccccccc} K & \xrightarrow{} & 0\\ \small {u}\big\downarrow & & \big\downarrow\small {} & \\ X & \xrightarrow{f} & Y \end{array}$
Suppose $f:X\to Y$ is a linear map between vector spaces $X$ and $Y$ in $\textbf{Vect}.$ Let $K=\{x\in X\mid f(x)=0\},$ and $u:K\to X$ be a linear map such that $f\cdot u=0$, hence $u\in \text{ker}(f).$ Let $K'=\{x\in X\mid f(x)=0\}$ and $u':K'\to X$ be another linear map such that $f\cdot u'=0,$ and so $f\cdot u=f\cdot u'=0.$ If $\theta:K'\to K$ is a linear map such that $u\cdot \theta=u'.$ Then $(u\cdot \theta)(x)=u(\theta(x))=u'(x),$ for all $x\in X,$ which implies $\theta(x)=x.$ If $\theta'$ is also another linear map such that $u\cdot \theta'=u',$ then $u\cdot \theta=u\cdot \theta'=u'.$ But the map $u:K\to X$ is an inclusion map, hence it is one-to-one. So by $\textbf{(3)(ii)},$ $u$ is a monomorphism, and so $\theta=\theta',$ which shows that $\theta$ is unique.
$\color{Red}{Questions:}$
I would like to know if I did the exercise correctly. I tried to imitate how Rotman did the similar exercise for modules. I have also attached screenshots of the diagrams from Rotman's text to make his proof easier to follow. I gave more details on showing the existence of $\theta:K'\to K.$ I am not sure why Rotman did not included it. Thank you in advance.
I have some trouble following your solution. What the exercise asks you to do is to start with $K = \{x\in X\mid f(x)=0\}$ and the morphism $u: K\to X$ (which is just the inclusion $u(x) = x$) and to show that the diagram which you have drawn satsifies the correct universal property. This means your solution can not start with "let $u:K\to X$ be a map" because the map $u$ is already given and fixed.
What you need to show is the following: Given any other vector space $W$ and any two maps $g:W\to X$ and $h:W\to 0$ such that $f\circ g = 0\circ h$, there is a unique map $\theta:W\to K$ such that $u\circ \theta =g$ and $0\circ \theta = h$. This is less complicated in diagram form, so I have made a picture for you:
I hope this helps, universal properties can be confusing at first. Good luck and write a comment if you stay stuck! :)