The trick for question (1) is not to use determinants. To select a basis of a vector field, you do not need commutativity. As a consequence, all well-known formulas like $$\dim (V/U)=\dim V-\dim U$$ (if $V$ is finite-dimensional and $U\subset V$ a linear subspace) still hold. For linear maps $F\colon V\to W$, you still have $\mathrm{rk} F=\dim\mathrm{im} F=\dim V-\dim\ker F$.

More concretely, one can define the rank of an $m\times n$-matrix $A$ over a skew field $K$ as $\mathrm{rk} A=\dim(\mathrm{im} A)$. One can compute it (e.g. using the Gauss algorithm), just as one would over a commutative field. It does not change if one passes from $K$ to $L$. Commutativity is not used in any of the proofs.

So, start with a matrix $A$. Left multiplication by so-called elementary $m\times m$-matrices describes row operations. All elementary matrices $C$ are invertible in the sense that there is another elementary matrix $D$ satisfieing $CD=DC=E_m$ (the only commutativity properties ever used are $k\cdot k^{-1}=k^{-1}k=1$ and $1\cdot k=k\cdot 1=k$, which hold in division rings).

Assume that you arrive at a matrix of the form $$B=\begin{pmatrix}\cdots&0&1&*&&\cdots\\&&\cdots&&0&1&*\cdots\\&&&&&\ddots\end{pmatrix}\;.$$ Then it is an exercise to read of $\dim\mathrm{im}B$, $\dim\ker B$ etc., and to check that one gets the same numbers for $A$. Also, the whole computation is valid as well over $L\supset K$.

The trick for question (1) is not to use determinants. To select a basis of a vector field, you do not need commutativity. As a consequence, all well-known formulas like $$\dim (V/U)=\dim V-\dim U$$ (if $V$ is finite-dimensional and $U\subset V$ a linear subspace) still hold. For linear maps $F\colon V\to W$, you still have $\mathrm{im} F\cong V/\ker F$, so $\mathrm{rk} F=\dim\mathrm{im} F=\dim V-\dim\ker F$.

More concretely, one can define the rank of an $m\times n$-matrix $A$ over a skew field $K$ as $\mathrm{rk} A=\dim(\mathrm{im} A)$. One can compute it (e.g. using the Gauss algorithm), just as one would over a commutative field. It does not change if one passes from $K$ to $L$. Commutativity is not used in any of the proofs.

So, start with a matrix $A$. Left multiplication by so-called elementary $m\times m$-matrices describes row operations. All elementary matrices $C$ are invertible in the sense that there is another elementary matrix $D$ satisfieing $CD=DC=E_m$ (the only commutativity properties ever used are $k\cdot k^{-1}=k^{-1}k=1$ and $1\cdot k=k\cdot 1=k$, which hold in division rings).

Assume that you arrive at a matrix of the form $$B=CA\begin{pmatrix}\cdots&0&1&*&&\cdots\\&&\cdots&&0&1&*\cdots\\&&&&&\ddots\end{pmatrix}\;,$$ where $C$ is an (invertible) product of elementary matrices. Then it is an exercise to read of $\dim\mathrm{im}B$, $\dim\ker B$ etc., and to check that one gets the same numbers for $A$. Also, the whole computation is valid as well over $L\supset K$.

Edit Alternatively, you can multiply the matrix $B$ above with elementary $n\times n$-matrices from the right (this corresponds to column operatorations) until it becomes a block matrix $$Z=BD=CAD=\begin{pmatrix}E_k&0\\0&0\end{pmatrix}\;,$$ where $D$ is an (invertible) product of elementary matrices and $E_k$ is a unit matrix and $k$ is the rank. The rank is the dimension of the image both on column vectors (where $A$ acts from the left) and on row vectors (where $A$ acts from the right).

For a column $v\in K^n$, you get $Av=0$ if and only if $Z(D^{-1}v)=0$, so $\dim\ker A=\dim\ker Z$. Similarly, $\dim\mathrm{im}A=\dim\mathrm{im}Z$ on columns. If one applies $A$ on rows $\alpha$, then for example $$\beta=\alpha A\Longleftrightarrow \beta=\alpha A=(\alpha C^{-1}) (ZD^{-1})\;,$$ so again $\dim\mathrm{im}A=\dim\mathrm{im}Z$ and $\dim\ker A=\dim\ker Z$ on row vectors.