I'm trying to compute the fundamental group of the following space (call it $X$):
Here is my attempt. Define $A,B\subset X$ as:
Clearly $A$ is homotopy equivalent to the figure $8$. Let $\alpha:=[a\cdot c^ {-1}], \beta:=[c\cdot d^{-1}]\in\pi_1(A)$ (here $a\cdot c^{-1}$ is the concatenation of $a$ with $c$ in the opposite direction). That way, $\pi_1(A)=\langle \alpha,\beta\rangle$. Similarly, for $\gamma:=[a\cdot b^{-1}],\delta:=[b\cdot d^{-1}]\in\pi_1(B)$, we have $\pi_1(B)=\langle\gamma,\delta\rangle$.
Furthermore, $A\cap B$ is homotopy equivalent to $\mathbb{S}^1$ and clearly $\pi_1(A\cap B)=\langle [a\cdot d^{-1}]\rangle$. Since $\alpha\beta=[a\cdot d^{-1}]$ and $\gamma\delta=[a\cdot d^{-1}]$, we have: $$\pi_1(X)=\langle \alpha,\beta,\gamma,\delta\mid \alpha\beta=\gamma\delta\rangle$$
Is this correct?
That is correct. Note that $\pi_1(X)$ has a much simpler presentation: $$\pi_1(X)=\langle \alpha,\beta,\gamma,\delta\mid\delta=\alpha\beta\gamma^{-1}\rangle=\langle\alpha,\beta,\gamma\rangle=\mathbb F_3$$ This can also be seen by noting that the original space $X$ is homotopy equivalent to $S^1\vee S^1\vee S^1$ (a bouquet of three circles), and applying van-Kampen to see that $$\pi_1(X)=\pi_1(S^1)*\pi_1(S^1\vee S^1)=\mathbb Z*\mathbb F_2=\mathbb F_3.$$