$\newcommand{\rank}{\operatorname{rank}}$
So your conclusion is not incorrect, and there are two things I have to say regarding this.
First and foremost, the notion of orthogonality is extra structure on a vector space is extra structure; that is it intrinsically depends on endowing a vector space with an inner product. If you don't know the abstract definition of an inner product, it is a bilinear map $\langle \cdot ,\cdot \rangle:V\times V\rightarrow \mathbb F$ which is symmetric, non-degenerate, and positive definite. Positive definiteness implies that $\langle v,v\rangle>0$ (and also implies the non-degeneracy but I include this to pave the natural generalization to pseudo-euclidean inner products). In the most common case of real vector spaces, we kind of take this extra structure for granted, because $\mathbb R^n$ kind of comes endowed with an obvious basis, and an obvious inner product given by $\langle x,y\rangle=x_1y_1+\cdots+x_ny_n$. However, the existence of such an object does not come for free in the definition of a vector space; that is there is some choice we have to make (indeed their are many inner products on $\mathbb R^n$). Once we have an inner product, we can begin to talk about angles between vectors (such as orthogonality), and lengths of vectors, but not before we have one. The inner product is essentially what begins to take us from linear algebra to geometry.
Now, the big problem here is that there is not a 'natural' way that a finite field (or fields of characteristic not $0$) can satisfy the positive definite characteristic. That is, there is no natural way for $\langle v,v\rangle$ to be greater than zero if the underlying field has characteristic $p$. Indeed, consider your $\mathbb F_4$, then if we naively apply the definition of the order on $\mathbb R$, we get that $0<1$ if $0+a<1+a$ for all $a$, but then $0+1<1+1=0$, so this doesn't make much sense. Instead, what we can do is relax the positive definiteness condition, and instead examine psuedo-euclidean inner products, which are bilinear maps $V\times V\rightarrow \mathbb F$ that are bilinear and non-degenerate. Note that the `usual' inner product on $\mathbb F_2^2$ satisfies this, but with pseudo-euclidean inner products we get kind of weird results like nonzero vectors of length zero such as $(1,1)\in \mathbb F_2$. (Note that this actually has a ton of importance outside of finite fields. In fact, the Minkowski inner product on $\mathbb R^4$ given by $x\cdot y=x_1y_1+x_2y_2+x_3y_3-x_4y_4$ is very important in physics, and is the background object for the study of special relativity and quantum field theory.)
Secondly, the proof linked in your post makes use of the orthogonality with a standard inner product, and as you've noted this will not work if instead you have a pseudo-euclidean inner product. I believe the best you can say the rank nullity theorom (I don't think this depends on the characteristic of your field but could be wrong?), i.e. for every linear map $A:V\rightarrow V$ we have that:
$$\dim V=\dim \rank (A)+\dim \ker (A)$$
With this, you can choose a basis for $\ker(A)\subset V$, extend it to a basis of $V$, and get a very unnatural (i.e. we had to make some choices here) isomorphism:
$$V\cong \mathbb F^{\dim\rank(A)}\oplus \ker(A)$$
