(Easy direction)
Suppose $x_0,x_1$ lie in the same path component and let $\sigma$ be a path from $x_0$ to $x_1$. Then $\sigma(0)=x_0$ and $\sigma(1)=x_1$ and so $\partial_1(\sigma)=x_1-x_0$. It follows that $x_1 - x_0 \in\operatorname{im}\partial_1$.
(Hard direction)
Suppose $x_1 - x_0 \in \operatorname{im}\partial_1$. Then there exists a finite set of paths $\sigma_i \colon\mathbf{I}\to X$, $i \in \{1, \ldots, k\}$ and coefficients $a_i$ such that $\partial_1 \sum a_i\sigma_i = x_1 - x_0$.
So, $$\begin{align}\sum a_i\sigma_i(1)-a_i\sigma_i(0) = x_1 - x_0\end{align}\tag{1}$$ by definition of $\partial_1$. Without loss of generality, we may assume that each of the $a_i$ is non-negative, as otherwise, replace $a_i$ with $-a_i$ and $\sigma(t)$ with $\sigma(1-t)$.
Let $A = \sum a_i$. If $A = 1$, then we have $\partial_1 \sigma = \sigma(1) - \sigma(0)$, so $\sigma(1) - \sigma(0) = x_1 - x_0$. By the fact that $S_0(X)$ is a free group generated by elements of $X$, we must have $\sigma(0) = x_0$ and $\sigma(1) = x_1$. So, $\sigma$ is a path from $x_0$ to $x_1$, hence they lie in the same path component. This is out base case.
Now in general, assume that $\sigma_1(0) = x_0$ (otherwise, reorder the $\sigma_i$s). Either $\sigma_1(1) = x_1$, in which case we're done, or else $\sigma_1(1) = v$ for some $v \neq x_1$. It follows that $\sum a_i \sigma_i(1)$ contains a positive $v$ component. So, there must be some $\sigma_j \neq \sigma_1$ such that $\sigma_j(0) = v$, as the right hand side of equation $(1)$ has no $v$ component. Without loss of generality, assume that $j=k$. Then, in the sum, we may replace one copy of $\sigma_0 + \sigma_k$ with their concatenation $\sigma_0\# \sigma_k$, as $\sigma_{k+1} = \sigma_0(1) = \sigma_k(0)$. This means we've replaced the sum $\partial_1 \sum_{i=1}^k a_i\sigma_i = x_1 - x_0$ with the sum
$$\partial_1 \left((a_1-1)\sigma_1 + (a_k-1) \sigma_k + \sigma_{k+1} + \sum_{i=2}^{k-1} a_i\sigma_i\right) = x_1 - x_0.$$
The new sum has $$A' = a_1-1 + a_k-1 +1 + \sum_{i=2}^{k-1} a_i\sigma_i = A-2+1 = A-1.$$
So we've reduced the sum of the coefficients by $1$. We can repeat the process until $A' =1$, in which case we've reached the base case and we're done.
Probably the important part to remember for the left to right implication is that free abelian groups have the nice property that if $n_1g_1 + \cdots n_k g_k = m_1h_1 + \cdots m_lh_l$ for integers $n_i,m_i$ and generators $g_i,h_i$, and the $n_i$ are all pairwise distinct, then $k=l$, the $m_i$ are all pairwise distinct also, and $g_i=h_j$ if and only if $n_i = m_j$.
