You can just use all the universal properties one at a time. We only need that $X \to \operatorname{Spec} k$ is qcqs and that $\Gamma(X,\mathcal O_X)$ is finite-dimensional; both these assumptions are satisfied if $X$ is proper over $k$. We also need that $T \to \operatorname{Spec} k$ is flat, which is always true if $k$ is a field.

Remark. Note that $\Gamma(T \times X, \mathcal O_{T \times X}) = \Gamma(T, \mathcal O_T) \otimes \Gamma(X, \mathcal O_X)$. Indeed, this follows from flat base change (Tag 02KH), since $T \to \operatorname{Spec} k$ is flat and $X \to \operatorname{Spec} k$ is qcqs.

Remark. If $V$ is a vector space and $W$ is a finite-dimensional vector space, then the natural map \begin{align*} V \otimes W &\to \operatorname{Hom}_k(W^\vee, V)\\ v \otimes w &\mapsto (\phi \mapsto \phi(w) v) \end{align*} is a natural isomorphism. For example, one can prove the case for $W$ of dimension $1$, and use that every $W$ decomposes as a finite direct sum of $1$-dimensional vector spaces. (Note, however, that this is false for $W$ infinite-dimensional, already if $V = k$.)

Lemma. Let $X$ be a qcqs $k$-scheme such that $\Gamma(X, \mathcal O_X)$ is finite-dimensional. Let $T$ be a $k$-scheme. Then $$\operatorname{Hom}_{\operatorname{Sch}/k}(T \times X, \mathbb A^1) = \operatorname{Hom}_{\operatorname{Sch}/k}(T, \operatorname{Spec} \operatorname{S}(\Gamma(X,\mathcal O_X)^\vee)).$$

Proof. By the adjunction $\operatorname{Ring}^{\operatorname{op}} \rightleftarrows \operatorname{Sch}$ given by $\Gamma$ and $\operatorname{Spec}$, we get $$\operatorname{Hom}_{\operatorname{Sch}/k}(T \times X, \mathbb A^1) = \operatorname{Hom}_k^{\operatorname{alg}}(k[x],\Gamma(T \times X, \mathcal O_{T \times X})).$$ By the universal property of $k[x]$, the latter is just $\Gamma(T \times X, \mathcal O_{T \times X})$. By the two remarks above, this equals $$\Gamma(T, \mathcal O_T) \otimes \Gamma(X, \mathcal O_X) = \operatorname{Hom}_k(\Gamma(X,\mathcal O_X)^\vee, \Gamma(T,\mathcal O_T)).$$ The universal property of the symmetric algebra turns this into $$\operatorname{Hom}_k^{\operatorname{alg}}(\operatorname{S}(\Gamma(X, \mathcal O_X)^\vee), \Gamma(T,\mathcal O_T)).$$ Finally, using the adjunction $\operatorname{Ring}^{\operatorname{op}} \rightleftarrows \operatorname{Sch}$ again, this finally becomes $$\operatorname{Hom}_{\operatorname{Sch}/k}(T, \operatorname{Spec} \operatorname{S}(\Gamma(X,\mathcal O_X)^\vee)),$$ which proves the claim. $\square$

You can just use all the universal properties one at a time. We only need that $X \to \operatorname{Spec} k$ is qcqs and that $\Gamma(X,\mathcal O_X)$ is finite-dimensional; both these assumptions are satisfied if $X$ is proper over $k$. We also need that $T \to \operatorname{Spec} k$ is flat, which is always true if $k$ is a field.

Remark. Note that $\Gamma(T \times X, \mathcal O_{T \times X}) = \Gamma(T, \mathcal O_T) \otimes \Gamma(X, \mathcal O_X)$. Indeed, this follows from flat base change (Tag 02KH), since $T \to \operatorname{Spec} k$ is flat and $X \to \operatorname{Spec} k$ is qcqs.

Remark. If $V$ is a vector space and $W$ is a finite-dimensional vector space, then the natural map \begin{align*} V \otimes W &\to \operatorname{Hom}_k(W^\vee, V)\\ v \otimes w &\mapsto (\phi \mapsto \phi(w) v) \end{align*} is a natural isomorphism. For example, one can prove the case for $W$ of dimension $1$, and use that every $W$ decomposes as a finite direct sum of $1$-dimensional vector spaces. (Note, however, that this is false for $W$ infinite-dimensional, already if $V = k$.)

Lemma. Let $X$ be a qcqs $k$-scheme such that $\Gamma(X, \mathcal O_X)$ is finite-dimensional. Let $T$ be a $k$-scheme. Then $$\operatorname{Hom}_{\textrm{Sch}/k}(T \times X, \mathbb A^1) = \operatorname{Hom}_{\textrm{Sch}/k}(T, \operatorname{Spec} \operatorname{S}(\Gamma(X,\mathcal O_X)^\vee)).$$

Proof. By the adjunction $\operatorname{Ring}^{\operatorname{op}} \rightleftarrows \operatorname{Sch}$ given by $\Gamma$ and $\operatorname{Spec}$, we get $$\operatorname{Hom}_{\textrm{Sch}/k}(T \times X, \mathbb A^1) = \operatorname{Hom}_k^{\operatorname{alg}}(k[x],\Gamma(T \times X, \mathcal O_{T \times X})).$$ By the universal property of $k[x]$, the latter is just $\Gamma(T \times X, \mathcal O_{T \times X})$. By the two remarks above, this equals $$\Gamma(T, \mathcal O_T) \otimes \Gamma(X, \mathcal O_X) = \operatorname{Hom}_k(\Gamma(X,\mathcal O_X)^\vee, \Gamma(T,\mathcal O_T)).$$ The universal property of the symmetric algebra turns this into $$\operatorname{Hom}_k^{\operatorname{alg}}(\operatorname{S}(\Gamma(X, \mathcal O_X)^\vee), \Gamma(T,\mathcal O_T)).$$ Finally, using the adjunction $\operatorname{Ring}^{\operatorname{op}} \rightleftarrows \operatorname{Sch}$ again, this finally becomes $$\operatorname{Hom}_{\textrm{Sch}/k}(T, \operatorname{Spec} \operatorname{S}(\Gamma(X,\mathcal O_X)^\vee)),$$ which proves the claim. $\square$