The following is taken from Fundamentals of Advanced Mathematics 1 : Categories, Algebraic Structures, Linear and Homological Algebra by Henri Bourlès.
Notations:
Let ${_\mathbf{K}}{\mathbf{Vec}}$ denote the category of left vector spaces over the division ring $\mathbf{K}$, $\operatorname{eq}(f,g)$ the equalizer of a pair of arrows $f,g$, and $\operatorname{coeq}(f,g)$ the coequalizer of a pair of arrows $f,g$.
Definitions for Equalizer and Coequalizer in ${_\mathbf{K}}{\mathbf{Vec}}$:
Equalizer
Let $\mathcal{C}={_\mathbf{K}}{\mathbf{Vec}}$ and $f:A\rightarrow B$ be a $\mathbf{K}$-linear function, we have $\operatorname{eq}(f,0)=\kappa:K\rightarrow A$ where $K=\ker(f):=f^{-1}(\{0\})$ is the definition of kernel of $f$ and $\kappa=\mathrm{can}$ is the canonical injection.
Coequalizer
Let $\mathcal{C}={_\mathbf{K}}{\mathbf{Vec}}$ and $f:A\rightarrow B$ be a $\mathbf{K}$-linear function, we have $\operatorname{coeq}(f,0)=\gamma:B\rightarrow C$ where $C=B/\operatorname{im}(f)$ is the usual definition of the cokernel $\operatorname{coker}(f)$ of $f$ (with $\operatorname{im}(f)=f(A)$, the image of $A$ under $f$) and $\gamma=\mathrm{can}$ is the canonical surjection.
I have few quick questions related to the above two definitions.
Does the notation ${_\mathbf{K}}{\mathbf{Vec}}$ denoting left vector space mean that it is a vector space with only left multiplications? I ask because of the following reasons, if we have $f,g:A\rightarrow B$ and $\kappa: K\rightarrow A$, then the equalizer of $f,g$ is $f\circ \kappa =g\circ \kappa,$ similarly, if we have again $f,g:A\rightarrow B$ and $\gamma:B\rightarrow C,$ then the coequalizer of $f,g$ is $\gamma \circ f=\gamma \circ g$. In the case of equalizer, isn't $\kappa$ being left multiplied by $f$ and $g$ and similarly in the case of coequalizer, isn't $\gamma$ being right multiplied by both $f$ and $g$.
This is more notational, what does the notation $0$ mean in $(f,0)$. We have $f,g:A\rightarrow B$, does $(f,0)$ mean $f:A\rightarrow B$ and $g=0:A\rightarrow B$? Also does $0:A\rightarrow B$ means that $g=0$ is either a zero map or a zero morphism?
The category of vector spaces over non-commutative division ring $K$ is just a category of, i this case, left $K$-modules. In particular, this category is abelian. Abelian categories have in particular all the features you ask for:
They are $\mathrm{Ab}$-enriched, meaning that each hom-set has a canonical structure of abelian group, satisfying all coherence conditions. In particular, each group of homomorphism has a zero element. Moreover, each equaliser or coequaliser of two parallel arrows is equivalent to (co-)equaliser of their difference and the zero morphism. In general, the equaliser of two arrows $f, g$ $$Eq \rightarrow A\rightrightarrows B$$ Is such a morphism that every morphism $h: X \rightarrow A$ making the diagram commute, i.e. such that $gh = fh$ factors uniquely through $Eq$. Its easy to see, that in case of modules over a ring, equaliser of a pair $(f, g)$ is the kernel of $f-g$, and equivalently, equaliser of a pair $(f-g, 0)$. Similar idea lies behind coequalisers, this time with all the arrows reversed - every map $h: X \rightarrow B$ such that $hf = hg$ factors through $Coeq(f, g)$. Again, a direct argument shows that the quotient map $B \rightarrow B/im(f-g)$ satisfies precisely the necessery and sufficient condition of an coequaliser.
Going back to your first confusion about multiplication - indeed, a left module by definition is equipped only with left multiplication. Structures considered in practise often have structure of bimodules, where the multiplication from both sides is considered, however the bimodule structure is just the same as a left module over the ring $R \times R^{op}$, since left actions can be canonically identified with right actions of the opposite ring, with the same elements but multiplication done in reversed order. In case of division algebras opposite ring $R^{op}$ is canonically isomorphic to $R$, which need not be the case in general. However, do not confuse multiplication by a scalar, which comes from the module structure, with composition of arrows - clearly arrows can be composed on both sides, whatever it means.
--- EDIT --- (more on kernels and cokernels)
The abstract nonsense definition of kernels and cokernels can be done as follows:
Given additive category $\mathcal C$, (by definition) the hom functor takes values in abelian groups. Defining the kernel of abelian group homomorphism as usual, we can then use it to define kernels in $\mathcal C$ using induced homomorphism on representable functors, which in the spirit of Yoneda representing the objects of $\mathcal C$. Formally, at first we construct the kernel functor $\operatorname{Ker}(f): \mathcal C^{op} \rightarrow \mathrm{Ab}$
$$ \operatorname{Ker}(f)(-) = \operatorname{ker}(\operatorname{Hom}(-, X) \xrightarrow{\circ f} \operatorname{Hom}(-, Y)) $$
After translating the problem to abelian groups by (enriched in abelian groups) representable functors, in the same way we go back, defining the kernel object $\operatorname{ker}(f)$ as representing kernel functor, so such that we have a natural isomorphism of functors
$$ \operatorname{Ker}(f)(-) \simeq \operatorname{Hom}(-, \operatorname{ker}(f)) $$
The definition may seem abstract, but it's just Yoneda lemma in enriched version applied twice to reduce the problem to category with already established solution. Similarly we treat cokernels, which are just kernels in the opposite category, i.e. $\operatorname{coker}(f) = (\operatorname{ker}(f^{op}))^{op}$
In this setting we can also define the image of $f: X \rightarrow Y$ as $$ \operatorname{im}(f) = \operatorname{ker}(Y \rightarrow \operatorname{coker}(f)) $$
and coimage as $$ \operatorname{coim}(f) = \operatorname{coker}(\operatorname{ker}(f) \rightarrow X) $$
One version of defining abelian categories is to require all the image of coimages to exist and being naturally isomorphic.
In case of categories such as modules or vector spaces, we easily see that these definitions agree with what we know - for example the cokernel being a quotient of $Y$ by the image means exactly that the image is the kernel of the quotient map.