Here's one possible approach. If a vector is in $\ker f$, then it will certainly be in $\ker(g\circ f)$, but a vector $v$ could also "survive" $f$ (that is $f(v)\neq0$, so $v\notin\ker(f)$), but have $f(v)\in\ker g$ so that still $g(f(v))=0$. The term $\dim(\operatorname{im}(f)\cap\ker g)$ in a sense measures how much this second possibility adds to the dimension of $\ker(g\circ f)$, because in that case $f(v)\in\operatorname{im}(f)\cap\ker g$.
To make the idea precise, one may note that for the purpose of determining $g\circ f$, one may replace $g$ by its restriction to $\operatorname{im}(f)$ (since any vector to which $g$ gets applied in the setting of $g\circ f$ lies in $\operatorname{im}(f)$)
$$
g\circ f = g|_{\operatorname{im}(f)}\circ f
$$
from which its follows that
$$
\operatorname{im}(g\circ f) = \operatorname{im}(g|_{\operatorname{im}(f)}).
$$
Now you can apply the rank-nullity theorem successively to $g\circ f$, to $g|_{\operatorname{im}(f)}$, and to $f$ to obtain the required identity of dimensions:
$$\begin{aligned}
\dim\ker(g\circ f)&=\dim U-\operatorname{rk}(g|_{\operatorname{im}(f)})\\
&=\dim U-(\dim\operatorname{im}(f)-\dim\ker(g|_{\operatorname{im}(f)}))\\
&=\dim\ker(f)+\dim(\ker(g)\cap\operatorname{im}(f))
\end{aligned}
$$
Another approach is to observe that the space $\ker(g)\cap\operatorname{im}(f)$ is precisely the image of $\ker(g\circ f)$ by $f$ (again the observations of the first paragraph apply; you can show both inclusions easily). Then apply the
rank-nullity theorem to the restriction of $f$ to $\ker(g\circ f)$.
