I apologize if this is extremely straightforward.
So I've been given a drawn graph of $f$, and it is asking me to draw the transformed graph $|f(-x)|$ and $\frac{1}{f(x)}$.
For $|f(-x)|$, I understand that the graph gets reflected over the y axis when it's $f(-x)$, but I don't quite understand what the absolute value signs do. Would I be correct in assuming that since the f(-x) can't be negative, all the values below the x axis would be reflected to above it?
I also don't completely understand how to do $\frac{1}{f(x)}$. I understand that the x intercepts of $f(x)$ would be the asymtotes of $\frac{1}{f(x)}$, but that aside, how is the graph supposed to look like compared to the original graph?
This problem is a very good test/builder of understanding of functions and graphs. You should try going back and forth between tables of $(x,y)$ pairs and graphs of the corresponding relation, equation or function. A few ones particularly come to mind: $y=|x|$ and $y=\frac1x$. Try graphing the former for $x\in\{-4,-3,-2,-1,0,1,2,3,4\}$, noting that $y$ has the same magnitude as $x$ but is a positive (or non-negative if $x=0$) sign. For the latter function, $y=\frac1x$, try plotting for (positive and negative) $x\in\{\pm\frac14,\pm\frac13,\pm\frac12,\pm1,\pm2,\pm3,\pm4\}$, noting that $y$ is always the reciprocal, which has the same sign as $x$.
Next, you can ask questions like, what happens to the graph of $y=f(x)$ when I replace $x$ by:
Lastly, note what happens when you do similar replacements with $y$. For example, take $y=|x|$ (which looks like $y=x$ in the first quadrant and $y=-x$ in the second quadrant, meeting in the point of a "V" at he origin). Replacing $x$ by $x-1$ shifts the graph to the right one unit, so the "V" has its point at $(1,0)$. However, replacing $y$ by $y-1$ gives the equation $y-1=|x|$, or, $y=|x|+1$, which shifts the graph up one unit to have its "vertex" point at $(0,1)$. Note that algebraic substitutions have an "indirect" effect, in the sense that $x$ has to compensate to play the same role in the equation (it has to be $1$ greater in my example), and similarly for $y$. So replacing either variable by itself minus one forces the variable's values to increase by one to be at the same place on the shape of the graph. This demonstrates the complementary nature of algebraic equations and their geometric graphs.
I hope this might help you getting started with reasoning about graphing.
So for your mystery function $f$ for which you have the graph, see if you can construct a table of values of $x$ (in the first colum) and corresponding values of $y=f(x)$ (in the second column). Start in the first row with $x=0$. Then find the value of the $y$-intercept, $f(0)$. Then try some values to the right and left of $0$ in further rows (Hint: take pairs of opposite values of $x$, such as $\pm1,\pm2$, etc). Next, add a column $|f(x)|$ and another column for $\frac1{f(x)}$, and fill in the values for these. Then try to graph these columns. Lastly, to graph $|f(-x)|$, flip the graph of $|f(x)|$ around the $y$-axis. For example, for $x=1$ take $y=|f(-1)|$, and for $x=-1$ take $y=|f(1)|$. Plot these points, and then figure out how the shape is transformed to inhabit these new points.
As became apparent in some comments, you can't really make approximate tables because you don't have scale on your graph. In this case, you could make up a scale and proceed, or sketch intuitively. For this, you have to understand the effect of the substitutions $x\to-x$ (mirror symmetry about the $y$-axis), $y\to|y|$ (reflects negative portions of the graph which are below the $x$-axis to their mirror-symmetric portions above it), and $y\to\frac1y$ (which keeps the same sign but inverts the magnitude of the height above or below the $x$-axis at each point). Here are some example sketches: