You can think about Rank-Nullity Theorem geometrically in terms of things called fibers over points.
Think about the case when your mapping $f: U \to V$ is surjective, and consider the mapping $f^{-1}: V \to 2^U$ that takes each point $p \in V$ to it's preimage $f^{-1}(p)$ (called fiber over $p$) in $U$. You can easily check the fibers are affine subspaces of $U$ parallel to each other (each point on $U$ passes through exactly one fiber). Also, the fiber passing through $0 \in U$ is exactly $\ker f$.
You can thus picture $U$ as being separated into infinite number of thin layers, like a sedimentary rock:

From this you can easily see that to uniquely specify a point in $U$ you can first specify a fiber (the set of fibers being parameterized by $V = \operatorname{im} f$) and then specify a point on a fiber (that is (non-uniquely) parameterized by $\ker f$). This gives you the Rank-Nullity Theorem: $\dim \ker f + \dim \operatorname{im} f = \dim U.$
For example, in case of a mapping $f: \mathbb{R}^2 \to \mathbb{R},\; (x, y) \to x + y$, the fibers will satisfy an equation of the form $y = a - x$ for some $a \in \mathbb{R}$. You can check that here $y + x = a - x + x = a$ indeed does not depend on either $y$ or $x$.
Now, how many independent variables do you need to specify a point in $\mathbb{R}^2$? You need one variable ($a$) to specify a fiber (equivalently, a point on $\mathbb{R}$), and another one (say, $x$) to specify a point on the fiber - that's two degrees of freedom, as expected!
The Rank-Nullity theorem states that for any surjective linear mapping $f: U \to V$, in any dimension you can use the same trick to uniquely parameterize any point in $U$. The same goes for any non-surjective linear mapping, of course, you'll just need to corestrict it to its image.
Alternatively, you could draw another line through $0 \in \mathbb{R}^2$ distinct from $\ker f$. You can easily show that it crosses each fiber of $f$ exactly once, so you can use it to parameterize fibers more explicitly: identify this line with $\operatorname{im} f$, then for any two points on $\ker f$ and $\operatorname{im} f$ you can uniquely obtain the corresponding point of $\mathbb{R}$ using the parallelogram rule. Rank-Nullity states that you can do this sort of thing in any dimension and for any $f$ (instead of lines you'll have affine subspaces of different dimensions, though).
This is a geometric picture of what's going on.