Let $K=\{a, b, c, ab, bc\}$ (in graph-theoretic terms, a path $a,ab,b,bc,c$, where $a, b,$ and $c$ are vertices and $ab$ and $bc$ are edges) be a simplicial complex and let $L=\{a, b, c\}$ be a subcomplex of $K$.
I have computed the first relative homology group, $H_1(K, L)=\ker \partial_1 / \text{im}~\partial_2 \simeq \mathbb{Z}^2$. Here, the boundary operator $\partial_n:C_n(K, L)\to C_{n-1}(K, L)$ is induced by the one defined on $C_n(X)$, where $C_n(K,L) := C_n(K)/C_n(L)$. Assume we are working with coefficients from $\mathbb{Z}_2$. Let $|K|:=\bigcup_{\sigma\in K} \sigma$ denote the union of the simplices of $K$ equipped with the topology from $\mathbb{R}^d$ for sufficiently large $d$. (Below assume $|L|\subseteq |K|$ is a subspace of $|K|$.)
By a theorem from Hatcher (Theorem 2.13 on page 114), $\overset{\sim}{H}_n(|K|, |L|)\cong \overset{\sim}{H}_n(|K|/|L|)$. This makes sense as if you quotient $|K|$ by $|L|$ you get two loops attached to the same point, which form the two "one-dimensional holes" that reflect in the first relative homology group being isomorphic to $\mathbb{Z}\times\mathbb{Z}$.
Is this a right way of thinking about relative homology?
You could use that isomorphism $H_n(|K|,|L|)\cong H_n(|K|/|L|)$ which holds for reduced homology, and then your analysis is correct. You could also interpret this more directly. Relative cycles are chains whose boundary lies in the subcomplex $L$. The two edges in $K$ satisfy this property, so give two relative cycles, which generate the $\mathbb Z\oplus\mathbb Z$ that you calculated.