Francesco, in comments, shows that any vector bundle on a curve degenerates to a direct sum of a line bundles. (By the way, I observe the convention that "degeneration" means moving towards the special fiber and "deformation" means moving away from it; you are doing degeneration.)

In the comments, the OP asks whether the moduli space of vector bundles on a curve with fixed rank and Chern character is connected. The answer is yes. Actually, there are a lot of subtleties in talking about this moduli space (stability issues), so I'll just directly answer the question about what can be connected to what in families over connected bases.

By Francesco's argument, we can degenerate from any vector bundle to a direct sum of line bundles. We need to show that, if $r=s$ and $\sum_{i=1}^r \deg L_i = \sum_{j=1}^s \deg M_j$, then we can build a path from $\bigoplus_{i=1}^r L_i$ to $\bigoplus_{j=1}^s M_j$.

**Step 1** On a curve, for any two ample line bundles $L_1$ and $L_2$, there is a degeneration from $L_1 \oplus L_2$ to $\mathcal{O} \oplus (L_1 \otimes L_2)$.

**Proof** Let $f_1$ and $f_2$ be sections of $L_1$ and $L_2$ with disjoint zero locus. Then
$$0 \to \mathcal{O} \stackrel{\begin{pmatrix} f \\ g \end{pmatrix}}{\longrightarrow} L_1 \oplus L_2 \stackrel{\begin{pmatrix} \otimes g & -f \otimes \end{pmatrix}}{\longrightarrow} L_1 \otimes L_2 \to 0$$
is exact. As in Francesco's argument, this shows we can degenerate from $L_1 \oplus L_2$ to $\mathcal{O} \oplus (L_1 \otimes L_2)$.

**Step 2** On a curve, for any $r$ line bundles $L_1$, $L_2$ ..., $L_r$, there is a path connecting $\bigoplus L_i$ to $\mathcal{O}^{\oplus(r-1)} \oplus \bigotimes L_i$

**Proof** Choose $D$ large enough that $\mathcal{O}(D)$ and $(\bigotimes L_i)(D)$ and all the $L_i(D)$ are ample. Recursively using Step 1 gives a family from $\bigoplus L_i(D)$ to $\mathcal{O}^{\oplus(r-1)} \oplus \bigotimes \left( L_i(D) \right)$. Step 1 also gives a path to this point from $\mathcal{O}(D)^{\oplus (r-1)} \oplus \left( \bigoplus L_i \right)(D)$.
So there is a path connecting $\mathcal{O}^{\oplus(r-1)} \oplus \bigotimes \left( L_i(D) \right)$ and $\mathcal{O}(D)^{\oplus (r-1)} \oplus \left( \bigoplus L_i \right)(D)$. Tensor that path with $\mathcal{O}(-D)$ to get a path joining $\bigoplus L_i$ to $\mathcal{O}^{\oplus(r-1)} \oplus \bigotimes L_i$.

So we can join $\bigoplus_{i=1}^r L_i$ to $\mathcal{O}^{\oplus (r-1)} \oplus \bigotimes L_i$ and we can do similarly with the $M$'s. If $\sum \deg L_i = \sum \deg M_i =d$ then, since $\mathrm{Pic}^d(X)$ is connected, we can find a path from $\bigotimes L_i$ to $\bigotimes M_i$.