As discussed in the comments, it is usually easiest to piece something like this together by hacking around than to follow a methodical approach. But I thought you might be interested in seeing a methodical approach written up.
In general, let's understand the Galois group $G$ of the splitting field of $x^4+bx^2+c$. Let the roots of $x^4+b x^2+c$ be $\pm \alpha$ and $\pm \beta$. We will assume that the polynomial doesn't have repeated roots. This is equivalent to $(b^2-4c)c \neq 0$.
Any Galois symmetry must either take the pair $\{ \alpha, -\alpha \}$ to itself, or to $\{ - \beta, \beta \}$, because these are the two two-element subsets of the roots which sum to $0$. So the group is a subgroup of the dihedral group $D_8$. I like to think of $D_8$ as the symmetries of a square, with $\pm \alpha$ and $\pm \beta$ at diagonally opposite corners of the square.
You act as if there are two four element subgroups of $D_8$, but there are really three: $C_4$, the copy of $V_4$ generated by reflections over lines parallel to the sides of the square and the copy of $V_4$ generated by reflections over the diagonals of the square. The last $V_4$ doesn't act transitively, so you rule it out at an earlier stage, but I'd rather keep it around.
Reflections over lines parallel to the sides of the square: Consider the element $\gamma: =\alpha \beta$ in the splitting field. If the full $D_8$ acts on the roots, then the orbit of $\gamma$ is $\pm \gamma$ and the stabilizer of $\gamma$ is this $V_4$. In general, if the group is $G$, then the stabilizer of $\alpha \beta$ is $G \cap V_4$. So $G$ is contained in this $V_4$ if and only if $\alpha \beta$ is fixed by the full Galois action, if and only if $\alpha \beta$ is rational.
Now, $(\alpha \beta)^2 = c$. So we get that $G$ is contained in this $V_4$ if and only if $c$ is square.
Reflections over the diagonals of the square: The element $\alpha^2$ is stabilized by this copy of $V_4$; so is the element $\beta^2$. So $G$ is contained in this $V_4$ if and only if $\alpha^2$ and $\beta^2$ are rational. Now, $\alpha^2$ and $\beta^2$ are the roots of $x^2+bx+c$, and the roots of this quadratic are rational if and only if $b^2-4c$ is square.
So $G$ is contained in this $V_4$ if and only if $b^2-4c$ is a square.
The group $C_4$: Again, I think of $C_4$ as a subgroup of the symmetries of a square -- specifically, the rotational symmetries. I am gong to find an element $\delta$ whose stabilizer is $C_4$; this will play a role analogous to $\gamma$ in the first section and $\alpha^2$ in the second.
I found $\gamma$ and $\alpha^2$ just by guessing, but $\delta$ took me a little thought.
I'd like a polynomial in $\alpha$ and $\beta$ which has odd degree in each, so that it is not fixed under reflection over any of the diagonals of the squares. We saw above that $\alpha \beta$ doesn't work -- it's stabilizer is $V_4$. Let's try a linear combination of $\alpha \beta^3$ and $\alpha^3 \beta$. A $90^{\circ}$ rotation of the square takes $\alpha \mapsto \beta$ and $\beta \mapsto - \alpha$, so it negates and switches the preceding monomials. In short, we take
$$\delta = \alpha \beta^3 - \alpha^3 \beta.$$
If all of $D_8$ acts, then the orbit of $\delta$ is $\pm \delta$. So, as above, the Galois group is contained in $C_4$ if and only if $\delta$ is rational.
Now,
$$\delta^2 = (\alpha^2 \beta^2) (\alpha^2 - \beta^2)^2 = c \cdot (b^2-4c).$$
(Remember that $\alpha^2$ and $\beta^2$ are the roots of $x^4+bx^2+c$.)
So $G \subseteq C_4$ if and only if $c(b^2-4c)$ is square.
In your case, we have $c \cdot (b^2 - 4c) = 2 \cdot (4^2-4 \cdot 2) = 16$, so your Galois group is contained in $C_4$.
By the way, notice that the intersection of any two of these groups is contained in the third. Correspondingly, if any two of $c$, $b^2-4c$ and $c \cdot (b^2-4c)$ are square, so is the third.
Ok, I finally got through it.
Let $K$ be the splitting field of $f$ over $F$. $K/F$ is Galois with degree $p$.
If $K$ lies in a radical extension $L$ of $F$. Then we have
$$
F=F_0\subset F_1\subset F_2\ldots\subset F_r=L
$$
where $F_i=F_{i-1}(\alpha_i)$ and $\alpha_i^{n_i}\in F_{i-i}$. We may assume that $\alpha_i\notin F_{i-1}$ and $n_i$ are all primes.
Let $K_i$ be $K(\alpha_1,\ldots\alpha_i)$, then $F_i \subset K_i$.
By induction, we can prove that $K_i/F_i$ is Galois with degree $p$ as follows.
First, $K_0/F_0$ is Galois with degree $p$.
We assume $K_{i-1}/F_{i-1}$ is Galois with degree $p$. $K_i=K_{i-1}(\alpha_i), F_i=F_{i-1}(\alpha_i)$. If $\alpha_i\in K_{i-1}$, then $F_i=F_{i-1}(\alpha_i)=K_i=K_{i-1}(\alpha_i)=K_{i-1}$ and $n_i=p$, since $[K_{i-1}:F_{i-1}]=p$. Because $\alpha_i\notin F_{i-1}$, $g=(t-\alpha_i)^p=t^p-\alpha_i^p$ is irreducible over $F_{i-1}$. Then the minimal polynomial of $\alpha_i$ over $F_{i-1}$ is $g$. However, $K_{i-1}/F_{i-1}$ is Galois, so $\alpha_i$ is separable, but $g=(t-\alpha_i)^p$, which shows that $\alpha_i$ is not separable.
This contradiction shows that $\alpha_i\notin K_{i-1}$. Note that $\alpha_i^{n_i}\in F_{i-1}\subset K_{i-1}$ and all $n_i$th roots of unity is in $F$. Then we have $g=t^{n_i}-\alpha_i^{n_i}$ is irreducible over $K_{i-1}$ and $[K_i:K_{i-1}]=n_i$. Then we can conclude that $K_i/F_i$ is Galois with degree $p$.
By induction, $K_i/F_i$ is Galois with degree $p$ for all $i$.
On the other hand, $K_r=F_r=L$, so $K_r/F_r$ is of degree 1, which leads to a contradiction.
Best Answer
Since $f(x) = x^4 - t$ is irreducible over $\mathbb{F}_p(t)$ and has degree $4$, if its splitting field has degree less than $8$ over $\mathbb{F}_p(t)$, it must have degree $4$.
Let $K = \mathbb{F}_p(t)(s)$, where $s^2 = -1$. A splitting field of $f$ over $\mathbb{F}_p(t)$ must contain $K$, so if the splitting field is of degree $4$ over $\mathbb{F}_p(t)$, then it must be of degree $2$ over $K$. In other words, $f$ must have non-trivial factors in $K[x]$. However, $f$ is irreducible over $K$, so this is not possible. Hence, the splitting field is of degree $8$ and is as you have computed.