Why is maltose a reducing sugar but not sucrose, even though they're both disaccharides?
The difference in stability between maltose and fructose boils down to the different structural elements their aldehyde/ketone groups have been turned into during the formation of the disaccharide.

In maltose, you combine two glucose units using the 1-hydroxy group of one and the 4-hydroxy group of the other. The 1-hydroxy group has been created in a process called acetal formation by the attack of what used to be the 5-hydroxy group and is now the ring oxygen. This is shown in equation $(1)$, the explicit carbon atom being carbon-1; the aldehyde.

$$\ce{R'-OH + RCHO &lt;=&gt;&gt; R'-OH^+-CHRO^- &lt;=&gt;&gt; R'-O-CHR-OH}\tag{1}$$

Notice that there are two oxygens bound to that carbon after the reaction; if we wanted, we could label it $\ce{RCH(OR)(OH)}$; this structure is called hemiacetal. If we use that hemiacetalic hydroxy group to bond to a second glucose unit,[1] we have replaced the hydrogen with an alkyl residue. Like the difference between alcohols ($\ce{R-OH}$) and ethers ($\ce{R-OR'}$), this group has a different reactitivy and is termed full acetal or O,O-acetal.

Hemiacetals are rather easy to break down as two mechanisms are available: you can either initially protonate the other oxygen or you can use a base to facilitate proton removal of the hydroxy group — ideally perhaps while offering the other oxygen a proton by another base. Full acetals, however, are not as easy to break down; the require protonation, implying a sufficiently strong acid.

Coming back to maltose, remember we have two glucose subunits. We used one to generate a full acetal but the other one (on the right-hand side of your scheme) is still hemiacetalic and can reconvert back to an aldehyde. This is why you preserve the reducing abilities; remember that equation $(1)$ is fully reversible and according to Le Châtelier removing aldehyde (e.g. by oxidation) will enhance the reverse reaction.

In sucrose, on the other hand, carbon 1 of glucose and carbon 2 of fructose are connected. These are the two carbon atoms that form hemiacetals in isolated glucose/fructose. Because of this, there is no hemiacetal group $\ce{RCH(OR)OH)}$ in sucrose — only full acetals — and no way it can easily revert back to an aldehyde or a ketone. Therefore, it has lost its reducing ability.

This is the key information of your quotation by the way. The anomeric carbon is the one that will be a hemiacetal. If two anomeric carbons are connected, they have formed a full acetal. Therefore, identifying the anomeric carbons and seeing if they still have a free hydroxy group (i.e., if they are hemiacetals) allows quick determination whether a disaccharide is reducing or not.



Note:

[1]: While formally correct; the actual reaction mechanism involves the loss of this hemiacetalic hydroxy group.