Since the 3-sphere S3 is a Lie group, the homotopy classes [X, S3] of a path-connected space X into S3 naturally form a group.
What is this group [T4, S3] if X is the 4-torus T4 ?
(Since T4 has a canonical parallelization, this group can be naturally viewed as the group of homotopy classes of non-singular vector fields on T4.)
Note that the $E_1$ structure on $S^3$ obtained from its Lie group structure agrees with the one coming from $S^3\simeq \Omega BS^3$. So all the structure maps in the Postnikov tower of $S^3$ deloop, and thus all the sequences below are compatible with the group structure.
We have a fiber sequence $\tau_{\geq 5} S^3 \to S^3 \to \tau_{\leq 4} S^3$. Mapping the $4$-torus into it tells us $[T, S^3]\cong [T,\tau_{\leq 4} S^3]$.
Next, we have a fiber sequence $$ \tau_{\leq 4} S^3 \to K(\mathbb{Z}, 3)\to K(\mathbb{Z}/2,5), $$ and mapping the torus into it gives an exact sequence $$ H^2(T,\mathbb{Z})\to H^4(T,\mathbb{Z}/2)\to [T, S^3]\to H^3(T,\mathbb{Z})\to 0. $$ As the k-invariant for $\tau_{\leq n+1}S^n$ for $n\geq 3$ is given by $\mathrm{Sq}^2$ (composed with mod 2 reduction), and this acts trivially on cohomology of the torus, the left hand map is zero. Furthermore, the right hand term is free, so the sequence splits, and $[T, S^3]\cong \mathbb{Z}^4 \oplus \mathbb{Z}/2$
To spell out what was indicated by user171227:
We can make use of the following facts:
Let $X$ be a based space.
(1) There is a weak equivalence $$ \Sigma (X^{\times n}) \simeq \bigvee_{k=1}^n \binom{n}{k} \Sigma X^{[k]}\, , $$ where $X^{[k]}$ is the $k$-fold smash power of $X$, and $jY$ means the wedge of $j$ copies of a based space $Y$.
(2) $\Sigma(X_+) \simeq (\Sigma X) \vee S^1$, where $X_+$ denotes $X$ considered as an unbased space with a disjoint basepoint.
Then we have $$ \Sigma (T^4_+) \simeq S^1 \vee 4 S^2 \vee 6 S^3 \vee 4 S^4 \vee S^5. $$ Therefore, \begin{align} [T^4,S^3] &\cong [\Sigma (T^4_+), BS^3]_\ast, \\ &\cong[\vee_{k=0}^4\tbinom{4}{k}S^{k+1},BS^3]_*,\\ &\cong [4S^4,BS^3]_\ast \times [S^5,BS^3]_\ast, \\&\cong [4S^3,S^3]_* \times [S^4,S^3]_\ast,\\ &=\pi_3(S^3)^{\oplus 4} \quad \oplus \quad \pi_4(S^3),\\ &\cong \Bbb Z^4 \oplus \Bbb Z/2, \end{align} where in the passage from line 2 to line 3 of the display we've used the fact that $BS^3$ is 3-connected.
Addendum: A similar argument shows that for any based space $Z$, there is an isomorphism $$ [T^n,\Omega Z] \cong \bigoplus_{k=0}^n \binom{n}{k}\pi_{k+1}(Z)\, . $$
Another way to do the computation is the via the Thom-Pontryagin construction. A the homotopy class of a map from $f:T^4 \to S^3$ is determined by the framed cobordism class of preimage of a regular value so a framed (and hence oriented) one manifold, $(\gamma_f, \Phi_f)$. The claim to check is this cobordism class is determined by the homology class of $\gamma_f$ in $H_1(T^4,\mathbb{Z})$ and a framing invariant in $\mathbb{Z}/2\mathbb{Z}$. Since the tangent bundle of $T^4$ is trivial (and indeed has a preferred trivialization (the Lie group framing) a framing of an oriented smoothly loop $\gamma$ in $T^4$ gives rise to a framing of $T(T^4)|_\gamma$ give by the normal framing the an oriented tangent vector field to $\gamma$, thus each component of $\gamma_f$ gives an element of $\pi_1(SO(4))=\mathbb{Z}/2\mathbb{Z}$. Note the choice here to define this framing obstruction is a choice of $Spin$-structure on $T^4$ (which you can think of as trivialization over the 2-skeleton of $T^4$. Thus the result has a pretty generalization (well known to Gauge theorists).
The components of the mapping space $[X^4,S^3]$ can be identified with
$\bullet$ $H_1(X,\mathbb{Z})\oplus \mathbb{Z}/2\mathbb{Z}$ if is $X$ is spin.
$\bullet$ simply $H_1(X,\mathbb{Z})$ is $X$ is not spin.
BTW there is a detailed discussion of this in Freed and Uhlenbeck's book, Instantons and Four-Manifolds. Look at Appendix B.
