The problem stated in the title comes from Vakil's AG notes 14.3.C. I am aware that there are few similar questions answered here. But I wish to give my own proof that I am not so sure if is valid. Firstly, let $\mathcal{F}$ be a locally free sheaf on $X=\mathbb{A}^1_k$ of rank $n$. Let me give a lemma that will be used.
Lemma (Geometry of Schemes, Corollary I-14). Let $\mathscr{U}$ be an open covering of a topological space $X$. If $\mathscr{F}_U$ is a sheaf on $U$ for each $U\in\mathscr{U}$, and if they agree on overlaps and satisfy cocycle conditions. There exists a unique sheaf $\mathscr{F}$ on $X$ such that $\mathscr{F}|_U\cong \mathscr{F}_U$ and such isomorphism respects inclusions.
Proof. Since $\mathcal{F}$ is locally free on $\mathbb{A}^1_k$, we see that it is quasi-coherent, which implies $\mathcal{F}\cong\widetilde{M}$ for some $k[x]$-module $M$. Assume $\{U_i\}$ is a trivialization cover of $\mathcal{F}$ over $X$. We can find some $\{D(f)\}$'s so that they cover these $U_i$. So one has $$\Gamma(D(f),\mathcal{F})\cong\Gamma(D(f),\widetilde{M})\cong\Gamma(D(f),\mathcal{O}_{\mathbb{A}^1_k}^n)\cong k[x]_f^n$$ We can extend these local information to a unique sheaf according to the lemma. Now our candidate is simply $\widetilde{k[x]^n}$, the uniqueness tells us this is exactly $\widetilde{M}$. Hence $$\mathcal{F}(X)\cong\widetilde{M}(X)=k[x]^n$$
Please give your critique freely. Thank you very much!