I don't think so: If $A$ is of finite representation type, the space $\mathrm{Ext}^1(M, N)$ is algebraically stratified so that each stratum corresponds to an isomorphism class (after base change to algebraic closure) of $X$. Consequently, there is a dense stratum in $\mathrm{Ext}^1(M, N)$ which gives a maximal extension (and it tends to have a minimal number of indecomposables, morally).

For example, take the quiver with vertices $I = \{1, \ldots, 4\}$ and with adjacency matrix
$$
Q = \begin{pmatrix}0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0\end{pmatrix}
$$
In other words, it is an oriented Dynkin quiver of type $D_4$ where the vertex $1$ is a sink and $3, 4$ are sources.
Let $M$ be the only indecomposable representation with dimension vector $(0, 1, 1, 1)$ and $N$ the indecomposable with dimension vector $(1, 1, 0, 0)$. For $i \in I$, let $P(i)$ denote the projective cover of the irreducible representation $L(i)$ corresponding to $i$.

Then $M$ has a projective resolution $0\to P(1)\oplus P(2)\to P(3)\oplus P(4)\to M\to 0$. Hence $\mathrm{Ext}^1(M, N)$ is of dimension $2$. We have the following two extensions of $M$ by $N$ who have $2$ indecomposables: $(1, 1, 1, 0)\oplus (0, 1, 0, 1)$ and $(1, 1, 0, 1)\oplus (0, 1, 1, 0)$. They represent two lines (minus the origin) in $\mathbf{C}^2$. However, there obviously exists an indecomposable extension of $M$ by $N$, which is the injective hull $I(1)$ of $L(1)$. In fact any element of the complement of $\mathbf{C}\times \{0\}\cup \{0\}\times\mathbf{C}$ is isomorphic to $I(1)$. Since this complement is Zariski-dense in $\mathbf{C}^2$, it follows that the largest number of indecompsables can be either $1$ or $2$, depending on the choice of basis.

I'm not expert of quiver representations, so please excuse me if the notations are terrible or if there is any error.