When does equality hold in the Minkowski's inequality $\|f+g\|_p\leq\|f\|_p+\|g\|_p$?

For $p = 1$, the proof uses the triangle inequality, $\lvert f(x) + g(x)\rvert \leqslant \lvert f(x)\rvert + \lvert g(x)\rvert$, and the monotonicity of the integral. You have equality $\lVert f+g\rVert_1 = \lVert f\rVert_1 + \lVert g\rVert_1$ if and only if you have equality $\lvert f(x) + g(x)\rvert = \lvert f(x)\rvert + \lvert g(x)\rvert$ almost everywhere. That means that almost everywhere at least one of the two functions attains the value $0$, or both values "point in the same direction", that is, have the same argument. You can simply formalise that as $f(x)\cdot\overline{g(x)} \geqslant 0$ almost everywhere.

For $1 < p < \infty$, in addition to the triangle inequality, the proof of Minkowski's inequality also uses Hölder's inequality,

$$\begin{align} \int \lvert f+g\rvert^p\,d\mu &\leqslant \int \lvert f\rvert\cdot\lvert f+g\rvert^{p-1}\,d\mu + \int \lvert g\rvert\cdot\lvert f+g\rvert^{p-1}\,d\mu\tag{1}\\ &\leqslant \lVert f\rVert_p \lVert f+g\rVert_p^{p-1} + \lVert g\rVert_p \lVert f+g\rVert_p^{p-1}.\tag{2} \end{align}$$

You then have equality $\lVert f+g\rVert_p = \lVert f\rVert_p + \lVert g\rVert_p$ if and only equality holds in $(1)$, and for both terms in $(2)$. Equality in $(1)$ is nearly the same as for the $p=1$ case, that gives a restriction $f(x)\cdot\overline{g(x)} \geqslant 0$ almost everywhere, except maybe on the set where $f(x)+g(x) = 0$ (but equality in Hölder's inequality forces $f(x) = 0$ and $g(x) = 0$ almost everywhere on that set, so at the end we really must have $f(x)\cdot \overline{g(x)} \geqslant 0$ almost everywhere if $\lVert f+g\rVert_p = \lVert f\rVert_p + \lVert g\rVert_p$). For equality of the terms in $(2)$, if $f+g = 0$ almost everywhere, we must have $f=g=0$ almost everywhere, and if $\lVert f+g\rVert_p > 0$, we have equality if and only there are constants $\alpha,\beta \geqslant 0$ with $\lvert f\rvert^p = \alpha \lvert f+g\rvert^p$, and $\lvert g\rvert^p = \beta\lvert f+g\rvert^p$ almost everywhere. We can of course take $p$-th roots, and together with the condition imposed by the triangle inequality, we obtain that

$$\lVert f+g\rVert_p = \lVert f\rVert_p + \lVert g\rVert_p$$

holds if and only if there are non-negative real constants $\alpha,\beta$, not both zero, such that $\alpha f(x) = \beta g(x)$ almost everywhere.

Let $q$ be a conjugate exponent of $p$, meaning $\frac{1}{q} + \frac{1}{p} = 1$. Now $||f + g||_p = ||f||_p + ||g||_p$ iff $||f+ g||_p^p = ||f + g||_p^{p - 1}(||f||_p + ||g||_p)$.

So since $||f + g||_p^p = \int_X |f + g|^p d\mu = \int_X (|f + g|)|f + g|^{p - 1} d\mu \leq \int_X (|f| + |g|)|f + g|^{p - 1} d\mu = \int_X |f||f + g|^{p-1} d\mu + \int_X |g||f + g|^{p-1} d\mu = ||f(f + g)^{p - 1}||_{L^1} + ||g(f + g)^{p - 1}||_{L^1} \leq ||f||_p||(f + g)^{p - 1}||_q + ||g||_p ||(f + g)^{p - 1}||_q = (||f||_p + ||g||_p)||(f + g)^{p - 1}||_q = (||f||_p + ||g||_p)(\int_X (|f(x) + g(x)|^{p - 1})^q d\mu)^{1/q} = (||f||_p + ||g||_p)(\int_X (|f(x) + g(x)|^{pq - q}) d\mu)^{1/q} = (||f||_p + ||g||_p)(\int_X (|f(x) + g(x)|^{p}) d\mu)^{1/q} = (||f||_p + ||g||_p)(\int_X (|f(x) + g(x)|^{p}) d\mu)^{1 - 1/p} = (||f||_p + ||g||_p)(\int_X (|f(x) + g(x)|^{p}) d\mu)^{\frac{p - 1}{p}} = ||f + g||_p^{p - 1}(||f||_p + ||g||_p)$ where the $3^{rd}$ inequality follows from the triangular inequality and the $6^{th}$ inequality holds from Holdor's inequality.

Therefore we have Minkowski's inequality gives us equality iff the triangular inequality is an equality almost everywhere and Holdor's inequality is an equality. Now the triangular inequality is an equality almost everywhere, iff $g(x)$ and $f(x)$ have the same sign almost everywhere, or if on every set of positive measure where the two differ by sign at least one of them is zero. We have equality for Holder’s if $|f|^p$ is a constant multiple of $|g|^q$ almost everywhere.