The case where $p=1$, the Minkowski's Inequality is nothing but a bunch of triangle inequalities of real numbers,
that is
$$\left( \sum_{K=1}^n |x_k + y_k|^p \right)^{\frac{1}{p}}
\le \left( \sum_{k=1}^n |x_k|^p\right)^{\frac{1}{p}} + \left( \sum_{k=1}^n |y_k|^p\right)^{\frac{1}{p}},
$$
when $p=1$, is
$$
\sum_{K=1}^n |x_k + y_k|
\le \sum_{k=1}^n |x_k| + \sum_{k=1}^n |y_k|.
$$
Equality holds when $x_ky_k\ge0$, for each k, that is $x_k$ and $y_k$ are either both non-negative or non-positive. So, that gives $x_k=c_ky_k$, for some positive $c_k$, for eack $k$.
For the case $p>1$ I assume you used
\begin{align*}
\left( \sum_{k=1}^n |x_k + y_k|^p \right)
& =\left( \sum_{k=1}^n |x_k + y_k|^{p-1}|x_k + y_k| \right) \\
& \le \sum_{k=1}^n |x_k + y_k|^{p-1}|x_k|+\sum_{k=1}^n |x_k + y_k|^{p-1}|y_k|,
\end{align*}
that is a triangle inequality, followed by Hölder's inequality:
$$
\sum_{k=1}^n |x_k + y_k|^{p-1}|x_k|
\le \left(\sum_{k=1}^n|x_k|^{p}\right)^{\frac{1}{p}}\left(\sum_{k=1}^n|x_k+y_k|^{(p-1)q}\right)^{\frac{1}{q}},
\quad \text{where } \frac{1}{p}+\frac{1}{q}=1.
$$
and similarly,
$$
\sum_{K=1}^n |x_k + y_k|^{p-1}|y_k|
\le \left(\sum_{k=1}^n|y_k|^{p}\right)^{\frac{1}{p}}\left(\sum_{k=1}^n|x_k+y_k|^{(p-1)q}\right)^{\frac{1}{q}}.
$$
and added the expressions to get the desired result.
So, to have equality in Minkowski's inequality we must have equality in both applications of Hölder's inequality and the triangle inequality, i.e., $\, \exists \, \lambda_1, \lambda_2 \ge 0$ s.t.,
\begin{align*}
|x_k|^p
=\lambda_1|x_k+y_k|^{(p-1)q}, \qquad
|y_k|^p
=\lambda_2|x_k+y_k|^{(p-1)q} \tag{1}
\end{align*}
and
\begin{align*}
|x_k+y_k|
=|x_k|+|y_k|, \tag{2}
\end{align*}
for $k = 1,2, \cdots, n$.
WLOG we may assume $\lambda_1,\lambda_2 > 0$. Therefore, from $(1)$ we have $|x_k|^p = \left(\frac{\lambda_1}{\lambda_2}\right)|y_k|^p$. Again for equality to hold in $(2)$ $x_k$ and $y_k$ must have same sign. So, that gives us the equality condition to be $x_k = cy_k$, for $k = 1,2, \cdots, n$ with $c = \left(\frac{\lambda_1}{\lambda_2}\right)^{1/p}$.
A direct proof (not via Hölder's inequality) is not only possible, but is also more natural, as it directly gets to the heart of the matter: convexity. It works the same for integrals and sums.
Let $f=af_0$ and $g=bg_0$ where $f_0,g_0$ are unit vectors.
Since the function $u\mapsto |u|^p$ is convex, for $t\in [0,1]$ we have
$$|tf_0(x)+(1-t)g_0(x)|^p\le t|f_0(x)|^p+(1-t)|g_0(x)|^p \tag1$$
pointwise. Integrate (or sum) both sides:
$$\|tf_0+(1-t)g_0\|_p^p\le t +(1-t) = 1 \tag2$$
With $t=\frac{a}{a+b}$ the above becomes
$$\|f+g\|_p^p\le (a+b)^p = (\|f\|_p+\|g\|_p)^p \tag3$$
which is the desired inequality.
Best Answer
There is no simple condition for equality here. For example if $f=\chi_{(0,3)},g=\chi_{(1,2)}$ then $||f+g||_\infty =||f||_\infty +||g||_\infty$.But f and g are not multiples of each other. I hope this example convinces you that one cannot write down simple N & S conditions for equality in Minkowski's inequality.