I am trying to determine the group that the set of matrices of the form $$ \begin{bmatrix} 1 & 0 & a \\ 0 & 1 & b \\ 0 & 0 & c \end{bmatrix} $$ with $a, b, c \in \mathbb{F},\, c \ne 0$ forms under multiplication. I should be getting a semidirect product $(\mathbb{F}^2,+) \rtimes \mathbb{F}^*$, but when you multiply $(a,b,c)$ by $(a',b',c')$, the terms don't line up. If $z, z'$ correspond to matrices $A, B$, then the only way I can get it to be a semidirect product is to have $z \circ z'$ correspond to $BA$, not $AB$. Is this expected?
To be more precise, $$ \begin{bmatrix} 1 & 0 & a' \\ 0 & 1 & b' \\ 0 & 0 & c' \end{bmatrix} \begin{bmatrix} 1 & 0 & a \\ 0 & 1 & b \\ 0 & 0 & c \end{bmatrix} = \begin{bmatrix} 1 & 0 & a+a'c \\ 0 & 1 & b+b'c \\ 0 & 0 & cc' \ \end{bmatrix}, $$ so letting $x = (a,b),\, y = c$ we get $(x,y) \circ (x',y') = (x+yx',yy')$, which means the homomorphsim $H \to \text{Aut}(N)$ with $H = F^*,\, N = (F^2, +)$ is given by $y \to yI_2$.
Also, for $F = F_2, F_3, F_4$ is there a way to remove the direct product? $F_2$ makes it boil down to $Z_2^2$ and for $F_3, F_4$ we get $Z_2 \rtimes Z_3^2,\, Z_3 \rtimes Z_2^4$ respectively, but I'm not sure whether this can simplify further.
Suppose $\theta: \mathbb{F}^* \rightarrow Aut(\mathbb{F}^2)$ is a homomorphism such that $\forall c \in \mathbb{F}^*, \theta(c)=\theta_c$, where $\theta_c((a,b))=(ca,cb)$ for all $(a,b) \in \mathbb{F}^2$.
You can check that $\theta_c \in Aut(\mathbb{F}^2)$.
Now consider the isomorphism $\sigma:(\mathbb{F}^2,+) \rtimes_{\theta} \mathbb{F}^* \rightarrow G$, where $G$ is the group of matrices you mentioned, such that $$\sigma(((a,b),c))=\begin{bmatrix} 1 & 0 & \frac{a}{c} \\ 0 & 1 & \frac{b}{c} \\ 0 & 0 & \frac{1}{c} \end{bmatrix}$$
We have
$$\sigma(((a',b'),c')((a,b),c))=\sigma((a',b')+\theta_{c'}((a,b)),c'c))=\sigma(((a',b')+(c'a,c'b),c'c))=\sigma(((a'+c'a,b'+c'b),c'c))=\begin{bmatrix} 1 & 0 & \frac{a'+c'a}{c'c} \\ 0 & 1 & \frac{b'+c'b}{c'c} \\ 0 & 0 & \frac{1}{c'c} \end{bmatrix}=\begin{bmatrix} 1 & 0 & \frac{a'}{c'} \\ 0 & 1 & \frac{b'}{c'} \\ 0 & 0 & \frac{1}{c'} \end{bmatrix}\begin{bmatrix} 1 & 0 & \frac{a}{c} \\ 0 & 1 & \frac{b}{c} \\ 0 & 0 & \frac{1}{c} \end{bmatrix}=\sigma(((a',b'),c'))\sigma(((a,b),c)).$$
You can also check that it is onto and one-to-one. Hence, $(\mathbb{F}^2,+) \rtimes_{\theta} \mathbb{F}^* \cong G$.