How to understand why an algorithm uses additive or multiplicative homomorphic
The Why question has a deep answer since it is deep research! The How question is a lot easier. To check that how an encryption scheme is homomorphic look at the general homomorphism;
Let $f:A\to B$ be a map that preserve the structures.
$$f(x \cdot y) = F(x)\cdot f(y)$$ then we say $f$ is homomorphism ( in a very simple terms).
This is a bit different in FHE settings since schemes are semantically secure ( most of them) so the comparison cannot be made on the range ( encrypted message space) so you need to decrypt to see that.
To check the addition ( or any operation that FHE ( or partial FHE) uses) take two different inputs $x_1 \neq x_2$ and apply the rule where $\mathcal{E}$ stands for FHE encryption and $\mathcal{D}$ stands for FHE decryption.
$$\mathcal{E}(x_1 + x_2) = \mathcal{E}x_2) \boxplus \mathcal{E}(x_1),$$ we cannot check this, so we need decryption.
$$x_1 + x_2 = \mathcal{D}\left(\mathcal{E}(x_2) \boxplus \mathcal{E}(x_2)\right)$$
As done for addition this can be done for any operation. Why $\boxplus$ instead of $+$, the operation on the encrypted messages doesn't need to be an addition for addition.
So the general rule, let $\oplus$ be the operation that you want to test its homomorphism. Then
$$x \oplus y = \mathcal{D}\left(\mathcal{E}(x) \boxplus \mathcal{E}(y)\right)$$ where $\boxplus$ can be equal to $\oplus$  or not.
For example in Goldwasser–Micali cryptosystem
\begin{align}
\mathcal{E}(b_1)\cdot \mathcal{E}(b_2) &amp;= x^{b_1} r_1^2 x^{b_2} r_2^2 \;\bmod\; n \\[6pt]
&amp;= x^{b_1+b_2} (r_1r_2)^2 \;\bmod\; n \\[6pt]
&amp;= \mathcal{E}(b_1 \oplus b_2).
\end{align}
the multiplication of the ciphertext is equal to the X-OR of the plaintext this is mostly used in fingerprint validation on encrypted data.
In the Paillier cryptosystem
\begin{align}
\mathcal{E}(m_1) \cdot \mathcal{E}(m_2) &amp;= (g^{m_1} r_1^n)(g^{m_2} r_2^n) \;\bmod\; n^2 \\[6pt]
&amp;= g^{m_1 + m_2} (r_1r_2)^n \;\bmod\; n^2 \\[6pt]
&amp;= \mathcal{E}(m_1 + m_2).
\end{align}
As we can see the multiplication of the ciphertext is equal to addition of the plaintext. This is used for the aggregation of the data, especially in databases.
The ( textbook) RSA cryptosystem  on the other hand
\begin{align}
\mathcal{E}(m_1) \cdot \mathcal{E}(m_2) &amp;= m_1^e m_2^e \;\bmod\; n \\[6pt]
&amp;= (m_1 m_2)^e \;\bmod\; n \\[6pt]
&amp;= \mathcal{E}(m_1 \cdot m_2)
\end{align}
uses the same operation on both plaintexts and ciphertexts.