For a sigma-finite measure space $(X,\Sigma,\mu)$, the weak $L^p$ (hereafter denoted $L^{p,\infty}$) is defined by
$$\|f\|_{L^{p,\infty}}:=\sup_{t>0}t\mu(|f|>t)^{1/p}, \qquad (1\leq p<\infty) \tag{1}$$
One can also show that this equals the infimum over all constants $C>0$ such that $\mu(|f|>t)\leq C^p/t^p$ for all $t>0$.
Let $1\leq p_{0}<p_{1}<\infty$ be fixed, and let $p=(t/p_{0})+(1-t)/p_{1}$ for $t\in (0,1)$. In an exercise in Reed and Simon Vol. 2, the reader is asked to show that $f\in L^{p,\infty}$ if and only if there exists a constant $C>0$, such that for every $\lambda>0$, there exists a decomposition $f=f_{0,\lambda}+f_{1,\lambda}$, where $f_{i,\lambda}\in L^{p_{i}}$ and
$$\left\|f_{i,\lambda}\right\|_{L^{p_{i}}}\leq C\lambda^{1-p/p_{i}}$$
for $i=0,1$. The reader is then asked to show that the infimum of all such $C>0$, which we denote by $C_{f}$ equals $\left\|f\right\|_{L^{p,\infty}}$.
The first part isn't very difficult and in fact my argument showed that
$$C_{f}\leq\max\left\{\left(\dfrac{p}{p-p_{0}}\right)^{1/p_{0}}\left\|f\right\|_{L^{p,\infty}}^{p/p_{0}},\left(\dfrac{p_{1}}{p_{1}-p}\right)^{1/p_{1}}\left\|f\right\|_{L^{p,\infty}}^{p/p_{1}}\right\}$$
and
$$\left\|f\right\|_{L^{p,\infty}}\leq\left((2C_{f})^{p_{0}}+(2C_{f})^{p_{1}}\right)^{1/p}$$
But I don't see why they should be equal--it seems like they omitted some exponents in the statement of the question. In fact, it seems that this is false as $C_{()}$ satisfies the triangle inequality, whereas $\left\|\cdot\right\|_{L^{p,\infty}}$ does not. If $f,g\in L^{p,\infty}$, then $f_{i,\lambda}+g_{i,\lambda}$ is a decomposition for $f+g$ satisfying
$$\|f_{i,\lambda}+g_{i,\lambda}\|_{L^{p_i}}\leq \|f_{i,\lambda} \|_{L^{p_i}}+\|g_{i,\lambda}\|_{L^{p_i}}\leq C_f \|f\|_{L^{p,\infty}}\lambda^{1-p/p_i}+C_g \|g\|_{L^{p,\infty}}\lambda^{1-p/p_i}$$
Whence, $C_{f+g}\leq C_f+C_g$.
Is there something I'm not seeing?
Indeed, these cannot be always equal: as you correctly observed, one is a norm and the other is not. And here is an explicit example: $f\equiv 1 $ on the unit interval $[0,1]$ with $p_0=2$, $p_1=4$ and $t=1/2$; subsequently $p=8/3$.
Clearly, $\|f\|_{L^{p,\infty}}=1$ for every $p$.
Given any $\lambda>0$, we can write $f=f_{0,\lambda}+f_{1 \lambda}$ with $$\left\|f_{i,\lambda}\right\|_{L^{p_{i}}}\leq \frac12\lambda^{1-p/p_{i}},\quad i=0,1$$ with both $f_{i,\lambda}$ constant. This is possible because $$ \frac12\lambda^{1-p/p_{0}}+\frac12\lambda^{1-p/p_{1}} = \frac12(\lambda^{-1/3}+\lambda^{1/3})\ge 1 $$