Let $n:\nabla^1\to I$ be the homeomorphism
$(1-t)e_0+te_1\mapsto t$. There exist a well defined function $\psi:\pi_1(X,x_0)\to H_1(X)$ where $[\sigma]\mapsto[\sigma\circ n]$.
Now consider the following diagram (sorry because it looks so ugly)
\begin{array}{ccccccccc} S^1 & \\ \uparrow{u(t)=e^{2\pi i t}} & \searrow{\sigma'\simeq \gamma'} & \\ I & \xrightarrow{\sigma\simeq \gamma} & X & \end{array}
We then have $\sigma_*':H_1(S^1)\to H_1(X)$ where
$cls(\sum m_i\delta_i)\mapsto \sum m_icls(\sigma\circ\delta_i)$.
And also
$[\sigma\circ n]=[\sigma'\circ u\circ n]=\sigma_*'[u\circ n]\in H_1(X)$
$[\gamma\circ n]=[\gamma'\circ u\circ n]=\gamma_*'[u\circ n]\in H_1(X)$
Please if someone could justify this equality $[\sigma\circ n]=[\sigma'\circ u\circ n]$ (I think it follows from the diagram because $\sigma'\circ u=\sigma$) and what exactly needs to be proved, I did not understand nothing in class.
Thank you so much in advance.
Let me first try to fill the notational gaps in your question.
$\nabla^1$ is certainly the standard $1$-simplex in $\mathbb R^2$. The elements of $H_1(X)$ are homology classes $[\sum m_i \delta_i]$ of chains $\sum m_i \delta_i$, where $\delta_i : \nabla^1 \to X$ are singular $1$-simplices and $m_i \in \mathbb Z$.
If you take a closed path $\sigma: I \to X$ based at $x_0$, you get an induced $\sigma' : S^1 \to X$ such that $\sigma' \circ u = \sigma$. This map induces the homomorphism $\sigma'_* : H_1(S^1) \to H_1(X)$.
By definition $\sigma \circ n = \sigma' \circ u \circ n : \nabla^1 \to X$. Also note $u \circ n : \nabla^1 \to S^1$. Thus by definition of induced maps $$\sigma'_*([u \circ n]) = [\sigma' \circ u \circ n] = [\sigma \circ n] .$$
The map $\psi : \pi_1(X,x_0) \to H_1(X)$ is now defined by $$\psi([\sigma]) = \sigma'_*([u \circ n]) .$$ It remains to show that $\psi$ is well-defined. That is, if we have a closed path $\gamma: I \to X$ based at $x_0$ which is path-homotopic to $\sigma$, then $\sigma'_*([u \circ n]) = \gamma'_*([u \circ n])$. But this is obvious: Since the paths $\sigma, \gamma$ are path-homotopic, the induced maps $\sigma', \gamma' : S^1 \to X$ are homotopic and therefore $\sigma'_* = \gamma'_*$.