Let $P$ be the partition of the interval $[a,b]$. If $P$ is divided into two partitions $P_1$ and $P_2$ such that $P_1$ covers the interval $[a,c]$ and $P_2$ covers the interval $[c,b]$, then $U(P, f, α) = U(P_1, f, α) + U(P_2, f, α)$. But if we take infimum of all such upper sums over different partitions on both sides there comes an inequality $$\inf\{U(P, f, α) : P\text{ is a partition of }[a,b]\} \ge \\\inf\{U(P_1, f, α) : P_1 \text{ is a partition of }[a,c]\} + \inf\{U(P_2, f, α) : P_2 \text{ is a partition of }[c,b]\}$$ Why?
Here α is a monotonic increasing function.
Actually, you have it backwards:
If $P_1$ and $P_2$ are obtained by $P_1 = \{t \in P : t \leq c\} \cup \{c\}$, $P_2 = \{c\}\cup \{t \in P : c \leq t\}$, then $$U(P, f, α) \ge U(P_1, f, α) + U(P_2, f, α)$$ With equality holding in particular when $c\in P$.
However, $$\inf \{U(P,f,\alpha) : P\text{ partitions }[a,b]\} =\\ \inf \{U(P,f,\alpha) : P\text{ partitions }[a,c]\} + \inf \{U(P,f,\alpha) : P\text{ partitions }[c,b]\}$$
These hold because if in a partition $P$ of $[a,b], c$ falls between two adjacent partition points $p_i < c < p_{i+1}$, then the correspoding partitions $P_1$ and $P_2$ include the intervals $[p_i,c]$ and $[c,p_{i+1}]$. All other intervals in $P_1$ and $P_2$ are also intervals in $P$, and the contributions to the upper sums from those intervals are the same. However, $U(P,f,\alpha)$ has the term $$(\max_{[p_i,p_{i+1}]} f)(\alpha(p_{i+1}) - \alpha(p_i))$$ where $U(P_1,f,\alpha) + U(P_1,f,\alpha)$ has the terms $$(\max_{[p_i,c]} f)(\alpha(c) - \alpha(p_i)) + (\max_{[c,p_{i+1}]} f)(\alpha(p_{i+1}) - \alpha(c))$$ If the two maximums have the same value, then that will also be the maximum over all $[p_i,p_{i+1}]$, and the sum will match the term from $U(P,f,\alpha)$. But there is no reason for the two maximums to be the same. The best we can say is $$\max_{[p_i,p_{i+1}]} f = \max\{\max_{[p_i,c]} f, \max_{[c,p_{i+1}]} f\}$$ so in general we only have $$U(P, f, α) \ge U(P_1, f, α) + U(P_2, f, α)$$
But if $c \in P$, then every interval of $P$ is in $P_1$ or $P_2$, and the sums are equal.
When we take the infimums over all partitions, we get the same inequality. But for any partition $P_1$ of $[a,c]$ and any partition $P_2$ of $[c,b]$, their union is a partition of $[a,b]$, having $c$ as one of the partition points. So $U(P_1 \cup P_2, f,\alpha) = U(P_1, f,\alpha) + U(P_2, f, \alpha)$. Thus $\{U(P,f,\alpha) : P\text{ partitions }[a,b]\}$ includes the sum of every pair of values from $\{U(P,f,\alpha) : P\text{ partitions }[a,c]\}$ and $\{U(P,f,\alpha) : P\text{ partitions }[c,b]\}$. So its infimum cannot be any higher than the sum of the infimums of the other two sets. Which means equality holds.