How to understand why an algorithm uses additive or multiplicative homomorphic If we take the example of LWE (Learning With Errors) how does one know if it's homomorphic by addition or multiplication?
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.