There are a couple of mistakes. First, the Reduced Row Echelon Form (RREF) of the matrix should be
\begin{equation}
\begin{pmatrix}
1 & 0 & 0 & -2 \\
0 & 1 & 0 & 3 \\
0 & 0 & 1 & 3
\end{pmatrix}
\end{equation}
It should be $-2$ in the corner, not $2$. But surprisingly, the basis for $\ker (\varphi)$ you found is actually correct (you must have made a second sign error in the computation which cancelled out with the first error).
and the dimension of the kernel $=1$. Since it's $\mathbb{R^4} \to \mathbb{R^3}$, the dimension of image has to be $3$.
This is correct.
We have pivots in the first $3$ colums, so we can say that $<(3,0,−1),(2,1,0),(0,−1,0)>$ is the image of $\varphi$. And for bases of $\varphi$, we can take $(3,0,−1),(2,1,0),(0,−1,0)$, as they are linearly independent.
This is an incorrect statement; this I think is more of a terminology mistake rather than a deep conceptual one. The proper statement is $\{(3,0,−1),(2,1,0),(0,−1,0)\}$ forms a basis for the image of $\varphi$. Recall that the image of a linear map is a subspace of $\mathbb{R^3}$, so it can't just consist of $3$ vectors, but a basis for a $3$-dimensional image consists of $3$ vectors.
By the way, for this particular example, there is a much easier way to determine a basis for the image of $\varphi$. You already mentioned that the image has dimension $3$. But notice that the target space $\mathbb{R^3}$ also has dimension $3$. Hence, $\text{image}(\varphi) = \mathbb{R^3}$. So there's a particularly obvious basis: $\{(1,0,0), (0,1,0), (0,0,1) \}$.
Lastly, to compute $[\varphi]_B^C$, the matrix of $\varphi$ with respect to the bases $B$ and $C$, what you have to do is for each vector $v \in B$, compute what $\varphi(v)$ is, and write it as a linear combination of vectors from $C$. The coefficients will then be the entries of the matrix
For example, the first vector in $B$ is $(1,0,0,0)$. So, now we have to evaluate $\varphi$ on this vector:
\begin{align}
\varphi(1,0,0,0) &= (1,0,-1) \\
&= (1,1,1) - (0,1,1) + (0,0,-1)
\end{align}
Notice that the coefficients are $1,-1,1$. So, the first column of $[\varphi]_B^C$ looks like
\begin{pmatrix}
1 & \cdot & \cdot & \cdot\\
-1 & \cdot & \cdot & \cdot \\
1 & \cdot & \cdot & \cdot
\end{pmatrix}
The second vector of $B$ is $(1,1,0,0)$. Now, we compute again:
\begin{align}
\varphi(1,1,0,0) &= (5,1,-1) \\
&= 5 (1,1,1) -4 (0,1,1) + 2(0,0,-1)
\end{align}
So, the first two out of four columns of $[\varphi]_B^C$ look like:
\begin{equation}
\begin{pmatrix}
1 & 5 & \cdot & \cdot\\
-1 & -4 & \cdot & \cdot \\
1 & 2 & \cdot & \cdot
\end{pmatrix}
\end{equation}
I'll leave it to you to figure out what the last two columns are (follow the same process I did).
Best Answer
Lets fix: $V:= \mathbb{R}^4, K:= \left\langle \begin{pmatrix}-1 \\ 0\\0\\1\end{pmatrix} ,\begin{pmatrix}1 \\ 3\\2\\0\end{pmatrix} \right\rangle , I:= \left\langle \begin{pmatrix}1 \\ 1\\1 \end{pmatrix} ,\begin{pmatrix} 0\\-2\\1 \end{pmatrix} \right\rangle , W:= \mathbb{R}^3$
Now clearly: $K \subset V$ and $I \subset W$, this means we have canonical maps: $\pi:V\twoheadrightarrow V/_K$ and $\iota:I \hookrightarrow W$ (the projection onto the quotient and the inclusion). Now by the dimension formula we know $\dim( V/_K)=2=\dim(I)$, hence there exists an isomorphism $\varphi:V/_K \to I$ (pick your favourite one).
Consider the morphism: $$\iota \circ \varphi \circ \pi: V\twoheadrightarrow V/_K \xrightarrow{\sim}I \hookrightarrow W.$$
Now since both, $\varphi$ and $\iota$ are monics, the kernel of $\iota \circ \varphi \circ \pi$ is the same as the kernel of $\pi$ which construction is $K$. Dually since $\varphi$ and $\pi$ are epics, the image of $\iota \circ \varphi \circ \pi$ is the same as the image of $\iota$ which by construction is $I$.
So $\iota \circ \varphi \circ \pi$ has the desired properties
Now a funfact at the end: by the homomorphism theorem any morphism with the desired properties factors in precisely that way and "only" depends on the choice of $\varphi$.