I'll answer Question 2 first. The answer is yes.

If $Z$ has more than two indecomposable summands, then $\text{soc}(Z)$ has at least three summands, but $X$ has simple socle since it's uniserial, so $Y$, the kernel of $Z\to X$ has at least two summands. But $Y$ is also uniserial, and so has simple socle.

For Question 1:

If $Z$ is indecomposable (uniserial), then it's easy to see what the exact sequence must be, and when it exists, so suppose $Z=Z_1\oplus Z_2$ is a direct sum of two indecomposables.

An indecomposable is determined by the list of it's composition factors from head to socle, which can be represented by a word $w$ in a set indexing the isomorphism classes of simples. Denote the module corresponding to a word $w$ by $[w]$, if there is such a module.

Since $Y$ has simple socle, its map to one of the $Z_i$, say $Z_1$, must be injective, but not surjective since the exact sequence doesn't split. But then its map to $Z_2$ must be surjective, or else $X$, the cokernel of $X\to Z$, has nonsimple head.

So if $X=[w]$ and $Y=[w']$ then the extensions all look like the following, with the obvious maps.

If $[ww']$ is a module, then we can take $Z=[ww']$

If $w=uv$ and $w'=vt$, where $[uvt]$ is a module, then $Z$ could be $[uvt]\oplus[v]$.

There could be more than one choice that works.

I'll answer Question 2 first. The answer is yes.

If $Z$ has more than two indecomposable summands, then $\text{soc}(Z)$ has at least three summands, but $X$ has simple socle since it's uniserial, so $Y$, the kernel of $Z\to X$, has at least two summands. But $Y$ is also uniserial, and so has simple socle.

For Question 1:

If $Z$ is indecomposable (uniserial), then it's easy to see what the exact sequence must be, and when it exists, so suppose $Z=Z_1\oplus Z_2$ is a direct sum of two indecomposables.

An indecomposable is determined by the list of its composition factors from head to socle, which can be represented by a word $w$ in a set indexing the isomorphism classes of simples. Denote the module corresponding to a word $w$ by $[w]$, if there is such a module.

Since $Y$ has simple socle, its map to one of the $Z_i$, say $Z_1$, must be injective, but not surjective since the exact sequence doesn't split. But then its map to $Z_2$ must be surjective, or else $X$, the cokernel of $X\to Z$, has nonsimple head.

So if $X=[w]$ and $Y=[w']$ then the extensions all look like the following, with the obvious maps.

If $[ww']$ is a module, then we can take $Z=[ww']$

If $w=uv$ and $w'=vt$, where $[uvt]$ is a module, then $Z$ could be $[uvt]\oplus[v]$.

There could be more than one choice that works.