Writing $1$ as a Linear Combination of Polynomials
No sane person would do that count: you can easily use Nullstellensatz to show that $(f_1,f_2,f_3)=\mathbb{C}[x,t]$.
Anyway, I'm insane: we have $$f_4=x(f_3-tf_1-4xf_2)-2f_2=4x^2+x+2$$ and $$f_5=x^2f_1-(tx+1)f_2=x^4-2x^2+1$$ Hence we have reduced the problem to one variable. $$x^2f_4-4f_5=x^3+10x^2-4$$ $$(4x^2-x)f_4-16f_5=4(x^2f_4-4f_5)-xf_4=39x^2-2x-16$$ $$(-4x^2+x+10)f_4+16f_5=x^2+12x+36$$ $$47f_6=4(-4x^2+x+10)f_4+64f_5-f_4=(-16x^2+4x+39)f_4+64f_5=47x+142$$ $$f_6=x+\frac{142}{47}$$ $$f_4(x)-f_4\left( -\frac{142}{47} \right)=\left(x+\frac{142}{47}\right) \left(4x-4 \frac{142}{47}+1\right)$$ $$1= f_4\left( -\frac{142}{47} \right)^{-1}\left[f_4 -\left(4x-4 \frac{142}{47}+1\right) f_6\right]$$ Now, if you want, you may substitute the expressions of $f_4,f_5,f_6$ in terms of $t_1,t_2,t_3$.
You may use $$f_1,f_2,f_3,\\tf_1, txf_1,t^2f_1, \\ xf_2,x^2f_2,t^2f_2, \\ xf_3, \text{ and }\, tf_3.$$
This will give you $11$ expressions of the form $$g_i(x,y)=l_i(t^2,x^2,tx,t^3,tx^2,x,t,tx^3,xt^3,t^4,t^2x^2,1),$$ where $l_i$ is a linear function. These actually are functions of $12$ arguments, but what you need is just to get rid of the terms which are not constant, and there are $11$ of those. Now, write the coefficients of each $g_i$ (not considering the constant) as a column, forming a $11\times11$ matrix $A$. What you need is to solve the linear system $A\lambda=0$ in order to find the real coefficients $\lambda_i$ which will get rid of all the nonconstant terms.
Edit.
If I made no mistakes, the matrix is
[ 1 0 0 0 0 0 0 -1 0 0 -2]
[ 1 0 0 0 0 -1 0 0 0 0 0]
[ 0 1 0 0 0 0 0 0 -2 0 0]
[ 0 0 1 0 1 0 0 0 0 0 0]
[ 0 0 5 1 1 0 0 0 0 0 0]
[ 0 0 0 -1 0 0 1 0 0 0 0]
[ 0 0 0 0 -2 0 0 0 0 1 0]
[ 0 0 0 0 0 1 1 0 1 0 0]
[ 0 0 0 0 0 0 1 1 1 0 0]
[ 0 0 0 0 0 0 0 0 0 1 1]
[ 0 0 0 0 0 0 0 0 0 5 1].
Edit.
However, I didn't get the result I expected. The only nontrivial solution up to scalar gets rid of the constant term too. $$f_1-2f_2+x^2f_2+t^2f_2-txf_1=0.$$
So this is of no use. Maybe some modification of this procedure can give you the result you are looking for.