A one form $\alpha: \mathbb{R}^n \to \mathbb{R}$ is given by \begin{equation} \alpha = f_1(x_1,...,x_n)dx_1 + \cdots f_n(x_1,...,x_n) dx_n \end{equation}
I don't understand how this gives an output in $\mathbb{R}$ since all the terms in $\alpha$ have $dx_i$ in them. Futhermore, is the kernel of $\alpha$ the set of points $(x_1,...,x_n)$ such that $f_1(x_1,...,x_n) = \cdots = f_n(x_1,...,x_n) = 0$?
A differential form on $\mathbb{R}^n$ can be described in the way you said. As inputs, differential forms take vector fields. We write vector fields in the form $$X=\sum_iX^i\frac{\partial}{\partial x^i}$$ for some smooth functions $X^i:\mathbb{R}^n\to\mathbb{R}$. The partial derivatives $\{\partial_i\}$ form a basis for the tangent space $T_x\mathbb{R}^n=\mathbb{R}^{n}$ at each point $x\in\mathbb{R}^n$. The way differential forms and vector fields interact, is that they are dual to each other, in the sense that $\{\partial_i\}$ and $dx^i$ are dual bases to one another. What this means is that $$dx^i(\partial_j)=\delta^i_j$$ This means that if you give $k$ vector fields $\{X_1,\dots,X_k\}$ to a differential $k$-form $\omega$, then you get $$\omega(X_1,\dots,X_k):\mathbb{R}^n\to\mathbb{R}$$ Specifying to $1$-forms, as per your question, this means that a differential $1$-form gives us a smooth function, once we provide it with a vector field $X$: $$\alpha(X)=\sum_if_iX^i:\mathbb{R}^n\to\mathbb{R}$$ I've written $\alpha=\sum_if_idx^i$ as you did.
If you are doing multivariable calculus, perhaps you have not yet seen vector fields written as $X=\sum_iX^i\partial_i$, but rather as functions $X:\mathbb{R}^n\to\mathbb{R}^n$. In this case, the components of the function $X=(X^1,\dots,X^n)$ correspond to the components of $X=\sum_iX^i\partial_i$.
Finally, the kernel of $\alpha$ is probably not what you want to consider. Instead, you would typically consider the kernel of $\alpha_x$, i.e. $\alpha$ at the point $x$. Then we can view $$\alpha_x:T_x\mathbb{R}^n=\mathbb{R}^n\to\mathbb{R}$$ as a linear map, and talk about its kernel in this context.
Given the question, you may not be familiar with the terminology I'm about to use - but there is a context in which it does make sense to talk about the kernel of $\alpha$. Namely, let $R=C^\infty(\mathbb{R}^n)$ denote the ring of smooth functions on $\mathbb{R}^n$. Then the vector space of vector fields $\Gamma(T\mathbb{R}^n)\cong C^\infty(\mathbb{R}^n,\mathbb{R}^n)=R^n$ is an $R$-module. A $1$-form $\alpha$ gives you an $R$-linear map $$\alpha:R^n\to R$$ You could consider the kernel of this linear map $\alpha$, if you wish. But its kernel will never be a set of points of the space $\mathbb{R}^n$.