Yes, transitive extension and transitive closure are different notions. If you start from a relation $R$ on some set, then the transitive extension $R_1$ of $R$ is obtained by adding to $R$ a pair $(a,c)$ whenever $(a,b)$ and $(b,c)$ are in $R$ for some $b$. Note that the transitive extension is in general not-transitive, while transitive closures are. For example, consider the two matrices of your question, and let $(i,j)$ denote the place corresponding to row $i$ and column $j$. Let $R$ be the relation represented by the first matrix, $R_1$ the relation represented by the second one. Then we say that $i$ is related to $j$ under $R$ if there is $1$ in place $(i,j)$ of the matrix representing $R$, we say they are not related if there is $0$. The same for $R_1$. In the first matrix, you have $1$ at place $(1,2)$, and $1$ in place $(2,3)$, meaning that $1$ is related to $2$ and $2$ to $3$. In the matrix, you have $0$ at place $(1,3)$, meaning that $R$ is not transitive, but you have $1$ in matrix $R_1$, which represents the transitive extension of $R$. Still $R_1$ is not transitive, because for example, using the same notation, $(1,2)$ and $(2, 4)$ are in $R_1$, but $(1,4)$ is not.
To obtain a transitive relation starting from a general relation $R$, you have to build its transitive closure, which is the least relation containing $R$ which is transitive, or equivalently, the set intersection of all transitive relations containing $R$.
You can relate the two notions by observing that, if you denote by $R_1$ the transitive extension of $R$, by $R_2$ the transitive extension of $R_1,\ldots,$ by $R_{i+1}$ the transitive extension of $R_i\ldots$, then the transitive closure of $R$ is the set union of all $R_i$'s.
Your proof can be streamlined. You suppose $\mathcal{R}^{+} \setminus \mathcal{R} \neq \emptyset$. Then you use that $\mathcal{R}^{+}$ is the smallest transitive relation containing $\mathcal{R}$, but this is not the case because $\mathcal{R}$ is transitive and $\mathcal{R}^{+} \not\subseteq \mathcal{R}$. So the whole case distinction is not needed and the construction using $\mathcal{F}$ is not needed.
If you use the fact that the transitive closure of $\mathcal{R}$ is equal to the intersection of all transitive relations containing $\mathcal{R}$, then there is an alternative proof.
Define the set $Tr=\{\text{transitive relation }\mathcal{S}\text{ over }A \mid \mathcal{R} \subseteq \mathcal{S}\}$.
Our above fact above becomes $\mathcal{R}^{+}=\bigcap_{\mathcal{S} \in Tr} \mathcal{S}$.
$\Rightarrow$
Since $\mathcal{R}$ is transitive and it contains $\mathcal{R}$, we know that $\mathcal{R} \in Tr$, hence $\mathcal{R}^{+} \subseteq \mathcal{R}$. Because $\mathcal{R} \subseteq \mathcal{S}$ for all $\mathcal{S} \in Tr$, we have that $\mathcal{R} \subseteq \mathcal{R}^{+}$.
$\Leftarrow$
Since $\mathcal{R}=\mathcal{R}^{+}$ and $\mathcal{R}^{+}$ is transitive, so is $\mathcal{R}$. $\quad\blacksquare$
For completeness, a proof of the above fact (using the definition of transitive closure as the smallest transitive relation containing the original)
Let $\mathcal{R}^{+}$ the smallest transitive relation, and let $\mathcal{R}^{\cap}$ be the intersection of all transitive relations containing $\mathcal{R}$. By definition of smallest, $\mathcal{R}^{+} \subseteq \mathcal{R}^{\cap}$. Now, since $\mathcal{R} \subseteq \mathcal{R}^{+}$ and $\mathcal{R}^{+}$ is transitive, $\mathcal{R}^{+}$ is one of the sets in the intersection of $\mathcal{R}^{\cap}$; hence $\mathcal{R}^{\cap} \subseteq \mathcal{R}^{+}$. $\quad \blacksquare$
This proof uses that the smallest transitive relation exists. If there is not a smallest transitive relation, then there are multiple minimal transitive relations (because the power set lattice is complete). You can take the intersection of these minimal elements to get a smaller transitive relation, thus getting a contradiction.
Best Answer
You are looking for transitive reductions. If $R$ is a partial order with finite intervals, you get exactly the covering relations that uniquely define the partial order via its Hasse diagram.