Concept in Simple Terms:

Imagine oxidation as "removing hydrogen" from the molecule. A 1° alcohol (R-$CH_{2}$-OH) has TWO hydrogens on the carbon bearing -OH. Removing one H gives an aldehyde (R-CHO), which still has ONE hydrogen left. Removing that last H gives the carboxylic acid (R-COOH).

Why does PCC stop? PCC (pyridinium chlorochromate) works in anhydrous dichloromethane. For an aldehyde to be oxidized further to a carboxylic acid, water must attack the carbonyl first (forming a gem-diol hydrate), and THEN the hydrate is oxidized. Since PCC is used in anhydrous conditions, no water is available to form the gem-diol intermediate, so the reaction halts at the aldehyde.

Why does $KMnO_{4}$ keep going? $KMnO_{4}$ is used in aqueous conditions. Water is present, so the aldehyde forms a gem-diol, which is then oxidized to the carboxylic acid. The reaction has enough energy and water to complete both steps.

Analogy: PCC is like a car that runs on dry fuel — once the fuel is gone (no water), it stops. $KMnO_{4}$ has a full tank plus a refueling station (water), so it keeps running until the final destination (carboxylic acid).