How to get a linear map with specified kernel and image
As long as $\dim W+\dim U=\dim V$ it is always possible to find such a map, since two linear spaces are isomorphic whenever they have the same dimension.
Suppose the equality holds, then $V/U\cong W$, and the composite map $T:V\to V/U\to W\to$ is always such that $\ker T=U$ and ${\rm im}\ T=W $. In this composite, $V\to V/U$ is the natural projection, $V/U\to W$ is the isomorphism, $W\to V$ is the inclusion.
This argument says $\dim W+\dim U=\dim V$ is sufficient for such $T$ to exist. The converse is obvious as you have observed. Hence we conclude:
Such $T$ exists $\iff\dim W+\dim U=\dim V$.
Some comment, since I sensed in your question that you are looking for some geometric insight into understanding linear spaces by studying linear maps on them: You may think this is too easy. Indeed it is. The reason is that linear spaces are isomorphic whenever they have the same dimension. And the isomorphism can be given by (when using matrix to represent a linear map) any invertible matrix. The problem is that they are not "natural", and hence these maps are usually not very meaningful and does not provide any insight of the structure of the spaces you are studying.
However, in practice, natural maps are favored. So your question would be more interesting if you impose some more conditions on $T$ or $U$ or $W$.
For example, consider this new question:
Can we find a linear map $T:V\to V$ such that $\ker T=U,\ {\rm im}\ T=W$, and $T(w)=w$ for all $w\in W$? In other words, $T$ annihilates $U$ while leaving $W$ intact.
Let's try to find a necessary condition for such $T$. Suppose it exists. Then
$\dim V=\dim U+\dim W$ as usual.
If $v\in U\cap W$, then on one hand, $T(v)=0$ because $v\in U$; while on the other hand, $T(v)=v$ because $v\in W$. Hence $v=0$. This means $U\cap W=0$ and $U+W$ is a direct sum $\implies\dim(U+W)=\dim U+\dim W=\dim V\implies U\oplus W=V$.
You see, by imposing one more condition on $T$, it is necessary that $U,W$ must be the direct summands of $V$, and not just that their dimensions add up to $\dim V$. Infact the converse is true (which could be an easy exercise). Hence
Such $T$ (with the extra requiment that $T(w)=w,\ \forall w\in W$) exists $\iff V=U\oplus W$.
A more insightful exercise is the following:
For a linear map $T:V\to V$, prove that $T^2=T\implies V=\ker T\oplus{\rm im}\ T$. Conversely, if $V=U\oplus W$, then there exists a linear map $T$ such that $T^2=T$, and $\ker T=U$, ${\rm im}\ T=W$.
This $T$ is "natural" because it represents a projection onto $W$, and this gives a geometric interpretation.
As you state yourself, if such a map exists then $V/U \cong W$. So let's assume this.
First define $f$ as the quotient map $$f: V \to V/U: v \mapsto [v]$$ which clearly has $U$ as its kernel. then define $i$ as the natural inclusion $$i:W \to V: v\mapsto v$$ which obviously has $W$ as its image.
Then compose them to get $$T: V\to V/U\cong W \to V$$ with the desired kernel and image.