The matrix A has eigenvalues $\lambda = \lambda_1, \lambda_2, \cdots , \lambda_n$ with the characteristic polynomial $$\lambda^n + a_{n-1}\lambda^{n-1}+\cdots+a_0=0$$ the eigenvalues of $At$ are $\lambda t$ and are solutions of $$(\lambda t)^n + (a_{n-1}t)(\lambda t)^{n-1}+\cdots+a_0 t^n=0$$
I do not understand why there is a $t$ after $a_{n-1}$, so why is it not $$(\lambda t)^n + a_{n-1}(\lambda t)^{n-1}+\cdots+a_0 t^n=0$$?
Your formulation is not very clear. The characteristic polynomial is a polynomial, so it involves an indeterminate that is usually written $x$ or $X$, not $\lambda$ which in your setup is a scalar (one particular eigenvalue).
What it seems your question is about is how the characteristic polynomial changes if we scale a matrix $A$ by a scalar $t$. One way to see this is to check the definition of the characteristic polynomial. Being the determinant of $XI_n-A$, every contribution is obtained by multiplying $n$ factors, each of which either comes from $XI_n$ or from $-A$. To get a term involving a power $X^k$, one needs $k$ factors from $XI_n$ and the remaining $n-k$ from $-A$; then scaling $A$ by $t$ results in scaling the coefficient of $X^k$ in the characteristic polynomial by $t^{n-k}$.
A similar argument can be made using the known effect of scaling of eigenvalues. If the characteristic polynomial of $A$ splits as $(X-\lambda_1)\ldots(X-\lambda_n)$ then the characteristic polynomial of $At$ splits as $(X-t\lambda_1)\ldots(X-t\lambda_n)$, and again we see that the coefficient of $X^k$ gets multiplied by $t^{n-k}$.
If $\lambda_1,\ldots,\lambda_n$ are the eigenvalues of $A$, then they are the roots of the characteristic polynomial $p_A(\lambda)$. So we can write
$$\begin{array}{rcl}p_A(\lambda) & = & (\lambda - \lambda_1)(\lambda - \lambda_2)\cdots(\lambda - \lambda_n) \\ & = & \lambda^n + a_{n-1}\lambda^{n-1} + \cdots + a_1\lambda + a_0\end{array}$$
If $v_i$ is the eigenvector of $A$ corresponding to $\lambda_i$, that is, $Av_i = \lambda_i v_i$ for any $i$, then $v_i$ is an eigenvector for $tA$ with eigenvalue $t\lambda_i$ since $tAv_i = t\lambda_i v_i$. So the eigenvalues of $tA$ are $t\lambda_1,\ldots,t\lambda_n$.
The eigenvalues of $tA$ are precisely the roots of the characteristic polynomial of $tA$, so we get
$$\begin{array}{rcl}p_{tA}(\lambda) & = & (\lambda - t\lambda_1)(\lambda - t\lambda_2)\cdots(\lambda - t\lambda_n) \\ & = & \lambda^n + (ta_{n-1})\lambda^{n-1} + \cdots + (t^{n-1}a_1)\lambda + t^na_0\end{array}$$
The two polynomials you've written don't have the right eigenvalues as roots. $$ (t\lambda)^n + (ta_{n-1})(t\lambda)^{n-1} + \cdots + (t^{n-1}a_1)t\lambda + t^na_0 = t^np_A(\lambda)$$ and so has roots $\lambda_1,\ldots,\lambda_n$, while the second has roots $t^{-1}\lambda_1,\ldots,t^{-1}\lambda_n$ as roots.
