For the kernel you must find a basis for the space of vector $\vec{x}$ satisfying
$$A\vec{x}=\vec{0}.$$
The image is all $\vec{y} \in \mathbb{R}[x]_3$ such that $A\vec{x}=\vec{y}$. So we are trying to find a basis for the space $$A\begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix}.$$ By matrix multiplication, this space is given by the linear combination of columns of $A$:
$$\vec{C}_1x_1+\vec{C}_2x_2+\vec{C}_3x_3+\vec{C}_4x_4,$$ where $C_n$ is the $n^{th}$ column of $A$.
Answer for Kernel:
To find the Kernel, we perform row reduction on $A$ to solve the system $A\vec{x}=\vec{0}$ and find that the resulting matrix $A'$ is given by $$A'=\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1\\ 0&0&0&0 \end{pmatrix}$$ Thus the kernel is given by $\text{Span} \{ \begin{pmatrix} 0 \\ 0 \\ 1 \\0 \end{pmatrix}\}$
Answer for Column Space:
Doing column operations is equivalent to doing row operations on $A^T$. To find the image, we row reduce $A^T$. Row reducing $A^T$ yields $$\begin{pmatrix} 1 & 0 & 1 & 0\\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 &1 \\ 0 & 0 & 0 & 0 \end{pmatrix}.$$ So the column space is given by $$\text{Span} \{ \begin{pmatrix} 1 \\ 0 \\ 1 \\0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ 0 \\1 \end{pmatrix} \} $$
There are two ways I see to attack the problem. First, there's the naive method, involving directly applying the definitions of kernel and image. Second, there's the matrix method, involving turning the linear operator into a matrix, then analysing the matrix using row reduction. Both methods are viable. You are using the naive method in the first part, and referring to the matrix method in the second part, but seem unable to figure out how to do it.
To continue the naive method for part 1, you have a system of two linear equations
\begin{align*}
-x_{11} - x_{21} + x_{12}+x_{22} &= 0 \\
x_{11} + x_{21} - x_{12} - x_{22} &= 0.
\end{align*}
Note that both equations are just negatives of each other. We can therefore let one of the variables depend on the other three, which can be free. So, in particular, if we let $t = x_{21}$, $s = x_{12}$, and $r = x_{22}$, then
$$x_{11} = s + r - t,$$
and hence
\begin{align*}
\begin{pmatrix} x_{11} & x_{12} \\ x_{21} & x_{22} \end{pmatrix} &= \begin{pmatrix} s + r - t & s \\ t & r \end{pmatrix} \\
&= s\begin{pmatrix} 1 & 1 \\ 0 & 0\end{pmatrix} + t\begin{pmatrix} -1 & 0 \\ 1 & 0\end{pmatrix} + r\begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix}.
\end{align*}
This gives you three vectors whose span contain the kernel of $\varphi$. If you verify that all three vectors are also in the kernel of $\varphi$, and that they are linearly independent, then they form a basis for the kernel.
The naive method can also be used for the second part. By your own calculation, an arbitrary transformed matrix takes the form,
\begin{align*}
\varphi(X) &= \begin{pmatrix} 0 & -x_{11} - x_{21} + x_{12}+x_{22} \\ x_{11} + x_{21} - x_{12} - x_{22} & 0 \end{pmatrix} \\
&= x_{11}\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} + x_{21}\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} + x_{12} \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} + x_{22}\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}.
\end{align*}
We now have a list of $4$ vectors that span the image of $\varphi$, but you might notice that they're very dependent. Reduce it down to a linearly independent list, and you'll have a basis.
If you want to do the matrix route properly, you'll need to find the matrix for $\varphi$ with respect to a basis (or two, but why complicate things?). It doesn't matter which basis you choose, so we'll pick an easy one:
$$\mathcal{B} = \left(\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}\right).$$
We now compute the matrix of $\phi$ with respect to this basis. Namely, we must transform each basis vector individually in turn, and write the resulting matrices as coordinate column vectors with respect to the given basis. These column vectors will form the columns of our matrix for $\varphi$.
Let $(e_{11}, e_{21}, e_{12}, e_{22}) = \mathcal{B})$. Then
\begin{align*}
\varphi(e_{11}) &= \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} = 0e_{11} + (-1)e_{21} + 1e_{12} + 0e_{22} \\
\varphi(e_{21}) &= \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} = 0e_{11} + (-1)e_{21} + 1e_{12} + 0e_{22} \\
\varphi(e_{11}) &= \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} = 0e_{11} + 1e_{12} + (-1)e_{12} + 0e_{22} \\
\varphi(e_{21}) &= \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} = 0e_{11} + 1e_{22} + (-1)e_{12} + 0e_{22}.
\end{align*}
We take the coordinates and write them as column vectors, and use them to form our matrix. Our matrix for $\phi$ becomes
$$A = \begin{pmatrix} 0 & 0 & 0 & 0 \\ -1 & -1 & 1 & 1 \\ 1 & 1 & -1 & -1 \\ 0 & 0 & 0 & 0 \end{pmatrix}.$$
The nullspace of this matrix will be the set of coordinate column vectors, with respect to $\mathcal{B}$, corresponding to vectors in the kernel of $\phi$. For example, the coordinate column vector
$$\begin{pmatrix} 0 \\ 1 \\ 1 \\ 0 \end{pmatrix}$$
belongs to the nullspace of $A$, because the matrix
$$0e_{11} + 1e_{21} + 1e_{12} + 0e_{22} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$$
belongs to the kernel of $\varphi$.
The columnspace of $A$ corresponds to the image of $\varphi$ in much the same way; it consists of coordinate column vectors with respect to $\mathcal{B}$, corresponding to vectors in the image of $\varphi$. If you reduce the columns down to a basis (there's a method involving row reduction), then they will correspond to a basis for the image of $\varphi$.
Best Answer
I'll be assuming your matrices are real, but nothing much would change if they were complex.
The first thing to note is that the the kernel is a subset of the domain and the image is a subset of the codomain - this trips a lot of people up. To check if a matrix is in the kernel, you apply $F$ to it (like you correctly suggest for matrix $b$). However to check if a matrix $A$ is an "image", you need to see if there exists a matrix $M$ such that $F(M)=A$. So matrix $a$ is not an image, because every matrix in the image of $F$ has zero in the upper right and lower left. Matrix $c$ is an image, since
$$ F\left(\begin{array}{cc}3 & 0\\ 0& -3\end{array}\right)=\left(\begin{array}{cc}3 & 0\\ 0& -3\end{array}\right)=c $$
To find bases for the kernel and image, you can start with a base for $M(2\times 2)$. The standard base is the four matrices:
$$ e_1=\left(\begin{array}{cc}1 & 0\\ 0& 0\end{array}\right)\quad e_2=\left(\begin{array}{cc}0 & 1\\ 0& 0\end{array}\right)\quad e_3=\left(\begin{array}{cc}0 & 0\\ 1& 0\end{array}\right)\quad e_4=\left(\begin{array}{cc}0 & 0\\ 0& 1\end{array}\right) $$
Then, you know that any matrix $A\in M(2\times 2)$ can be written as $$ A=ae_1+be_2+ce_3+de_4. $$ Now apply $F$ to your basis elements: $$ F(e_1)=e_1\quad F(e_2)=0\quad F(e_3)=0\quad F(e_4)=e_4 $$
Since $F$ is linear, we can describe its range by taking an arbitrary element of $M(2\times 2)$, writing in terms of our base $\{e_1,e_2,e_3,e_4\}$, and applying $F$ to it: $$ F(ae_1+be_2+ce_3+de_4)=aF(e_1)+bF(e_2)+cF(e_3)+dF(e_4)=ae_1+de_4 $$
Thus the image can be described using the two basis matrices $e_1$ and $e_4$:
$$ \text{Im}F=\{\alpha e_1+\beta e_4:\alpha,\beta\in\Bbb{R}\} $$
Like you suggest, the kernel can be described as any matrix with zero diagonal. Now that we have a basis for $M(2\times 2)$, we can write this succinctly as
$$ \text{Ker}F=\{\alpha e_2+\beta e_3:\alpha,\beta\in\Bbb{R}\} $$
You should check to make sure these are subspaces.
Bonus question: show that any matrix $A\in M(2\times 2)$ can be written as the sum of a matrix in the kernel of $F$ and a matrix in the image of $F$.