Let $X$ be a topological space such that $\pi_n(X)=H_n(X)=Z$. A continuous map $f: S^n \rightarrow X$ is an element of $\pi_N(X)=Z$ therefore $[f]_{\mathrm{homotopy}}$ is characterised by an integer $k$.
I am trying to show that $k$ is the degree of $f$. Is the following correct?
By Hurewicz theorem we have a group homomorphism $h:\pi_n(X)\rightarrow H_n(X)$. Let $[c]$ be a generator of $H_n(X)$, i.e. $[c]_{\mathrm{homology}}=1$. The Hurewicz map is \begin{equation} h([f]_{\mathrm{homotopy}})=f_*[c]_{\mathrm{homology}} .\end{equation} For $[f]\in\pi_n(X)$ on one hand we have, by definition of degree \begin{equation} h[f]=f_*[c]=\mathrm{deg}(f)\,[c]. \end{equation} On the other hand if $[f]=k$, since the Hurewicz map is a group homomorphism we also have \begin{equation} h[f]=h[k]=k\, h[1]=k\, [c]. \end{equation} It follows that the integer characterising the homotopy class of $f$ is the degree of $f$.
Here is a counterexample. Consider $X = \mathbb{R}P^3$, an orientable 3-manifold and so $H_3(\mathbb{R}P^3)=\mathbb{Z}$. The universal covering map $S^3 \to \mathbb{R}P^3$ induces an isomorphism on $\pi_3$ (by the long exact sequence of homotopy groups for a covering map). Since the identity map on $S^3$ represents a generator of $\pi_3(S_3)=\mathbb{Z}$ it follows that the universal covering map $S^3 \to \mathbb{R}P^3$ represents the generator of $\pi_3(\mathbb{R}P^3) = \mathbb{Z}$. But the universal covering map has degree 2.
In your proof, the expression $f_*[c]$ makes no sense, because the domain of $f_*$ is $\pi_n(S^n)$ or $H_n(S^n)$ (either one is OK because they are naturally isomorphic by the Hurewicz theorem) but $[c]$ is an element of $H_n(X)$.