Is there a name for the following categorical concept? If so, where can I read about it?
Given some category $\mathcal{C}$ we build a new category $\mathcal{C}^3$ whose objects are $3$-term sequences of objects with morphisms from $\mathcal{C}$ $X\to Y \to Z$ and where a morphism consists of morphisms $X\to X'$, $Y\to Y'$ and $Z\to Z'$ such that the following diagrams commute
This is a bit too long for a comment, but I'm not sure it really answers the question. It's an alternate construction that realizes your category as a full subcategory of the category of spans in $\mathcal{C}$.
The morphisms here look like morphisms in a category of spans.
In other words, this looks like the functor category $\newcommand\C{\mathcal{C}}\newcommand\span{{\cdot \leftarrow \cdot \to \cdot}}\C^\span$.
However, we really want the full subcategory, $\newcommand\D{\mathcal{D}}\D$, of spans $$Y\overset{f}{\leftarrow} X \xrightarrow{h} Z$$ such that $h$ factors through $f$, i.e., such that there is some map $g:Y\to Z$ with $h=gf$.
Now $\D$ is not isomorphic to the category in your question, since your category remembers the factorization of $h$. However, these two categories are equivalent.
We define a functor that sends the object $X\xrightarrow{f} Y\xrightarrow{g} Z$ to the span $$Y\overset{f}{\leftarrow} X\xrightarrow{gf}Z.$$ The morphisms in the two categories are the same, so this functor is fully faithful, and $\D$ is essentially by definition the image of this functor, so this functor is also surjective. Therefore this functor is an equivalence.
Comment
I'm not sure there is a great definition of your category (up to isomorphism) other than the one you've given, because your choice of morphisms causes the category to "forget" about $g:Y\to Z$ and only remember the composite morphism $h:X\to Z$. As long as $gf=g'f$, the two objects $$X\xrightarrow{f} Y\xrightarrow{g} Z$$ and $$X\xrightarrow{f} Y\xrightarrow{g'} Z$$ are isomorphic via the morphism $(1_X,1_Y,1_Z)$.