MY ATTEMPT
Let us assume that $\mathcal{R}\subset X^{2}$ is transitive, that is to say, given $(x,y)$ and $(y,z)$ in $\mathcal{R}$, then $(x,z)$ also belongs to $\mathcal{R}$. According to such property, one has \begin{align*} (x,z)\in\mathcal{R}\circ\mathcal{R} \Longrightarrow \exists y\,(y\in X)\wedge(x\mathcal{R}y)\wedge(y\mathcal{R}z)) \Longrightarrow (x,z)\in\mathcal{R}\Longrightarrow\mathcal{R}\circ\mathcal{R}\subset\mathcal{R}. \end{align*}
Now it is time to go backwards. Let us prove it by contradiction. Assume that $x\mathcal{R}y$ and $y\mathcal{R}z$, but $(x,z)\not\in\mathcal{R}$. According to the definition of composition of relations, we have \begin{align*} (x,z)\in\mathcal{R}\circ\mathcal{R}\subset\mathcal{R}\Longrightarrow (x,z)\in\mathcal{R} \end{align*} which is a contradiction, as proposed.
MY QUESTION
Can someone double check my solution and provide a geometrical interpretation for such result?
The formulation could be a little more careful, e.g. "$(x,z)\in\mathcal R$" in itself does not imply "$\mathcal R\circ\mathcal R\subset \mathcal R$", as your last arrow seems to say. Apart from that, the argument is correct.
Note that your argument does not make use of proof by contradiction: is not necessary to start with assuming $\mathcal R$ is not transitive (and in fact that assumption is not used in your argument). If you leave that assumption out, the whole proof still stands: $\mathcal R\circ \mathcal R\subset \mathcal R$ and $x\mathcal Ry$ and $y\mathcal Rz$ implies that $(x,z)\in\mathcal R\circ\mathcal R$, thus $x \mathcal Rz$ by inclusion and since $x,y,z$ were arbitrary we see that $\mathcal R$ is transitive.