$f$ is a function. So is $g\text{.}$ There is no problem with an element of the codomain not being the image of any input, and there is no problem with $a$ from the codomain being the image of both 2 and 3 from the domain. We could use our two-line notation to write these as

\begin{equation*} f= \begin{pmatrix} 1 \amp 2 \amp 3 \amp 4 \\ d \amp a \amp c \amp b \end{pmatrix} \qquad g = \begin{pmatrix} 1 \amp 2 \amp 3 \amp 4 \\ d \amp a \amp a \amp b \end{pmatrix}\text{.} \end{equation*}

However, $h$ is NOT a function. In fact, it fails for two reasons. First, the element 1 from the domain has not been mapped to any element from the codomain. Second, the element 2 from the domain has been mapped to more than one element from the codomain ($a$ and $c$). Note that either one of these problems is enough to make a rule not a function. In general, neither of the following mappings are functions:

$Three black dots in a row above and two white dots in a row below. An arrow points from the left black dot to the left white dot and from the right black dot to the right white dot.$

$Three black dots in a row above and four white dots in a row below. Two arrows point from the left black dot to each of the two left-most white dots. Each of the other black dots have an arrow pointing down to the white dots below and slightly to the right of them.$

It might also be helpful to think about how you would write the two-line notation for $h\text{.}$ We would have something like:

\begin{equation*} h=\twoline{1 \amp 2 \amp 3 \amp 4}{\amp a,c? \amp d \amp b}\text{.} \end{equation*}

There is nothing under 1 (bad) and we needed to put more than one thing under 2 (very bad). With a rule that is actually a function, the two-line notation will always “work”.