The controlled unitary gate $CU$, for some general unitary gate $U$, is (1):
$$CU=|0⟩⟨0|⊗I+|1⟩⟨1|⊗U$$
From the Wikipedia article on the CNOT gate (https://en.wikipedia.org/wiki/Controlled_NOT_gate) a general unitary gate can be expressed as $U=e^{iH}$, where $H$ is Hermitian, and $CU$ can be written as (2):
$${CU=e^{i{\frac {1}{2}}(I_{1}-Z_{1})H_{2}}}$$.
How can we prove expression $(2)$ from expression $(1)$?
I'm going to use the direct product symbol $\otimes$ and the ordering of matrices to represent which operator acts on which space, instead of numbering them with "1" and "2".
With that notation, let your matrix in your exponential be written as:
$$
M
=
i\frac{1}{2}\left(I - Z\right)\otimes H
$$
$$
= i |1\rangle\langle 1| \otimes H\;.
$$
Then
$$
M^2 = -|1\rangle\langle 1|\otimes H^2
$$
and
$$
M^3 = -i|1\rangle\langle 1|\otimes H^3\;,
$$
et cetera (since $|1\rangle\langle 1| = |1\rangle\langle 1|^2 = |1\rangle\langle 1|^3$, et cetera).
Thus
$$
e^{M} = I\otimes I + M +\frac{1}{2!}M^2+\ldots
$$
$$
= I\otimes I + |1\rangle\langle 1|\otimes\left(iH + \frac{1}{2!}(iH)^2 + \ldots\right)
$$
$$
=\left(|0\rangle\langle 0|+|1\rangle\langle 1|\right)\otimes I + |1\rangle\langle 1|\otimes \left(iH + \frac{1}{2!}(iH)^2 + \ldots\right)
$$
$$
=|1\rangle\langle 1|\otimes\left(I + iH + \frac{1}{2!}(iH)^2 + \ldots\right) + |0\rangle\langle 0|\otimes I
$$
$$
=|0\rangle\langle 0|\otimes I + |1\rangle\langle 1|\otimes e^{iH}
$$
$$
=CU
$$